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. 5077
PRINCIPLES OF RADIO
PRINCIPLES OF RADIO
BY
. KEITH HENNEY, M.A.
Associate Editor, Electronics
Member, Institute of Radio Engineers
NEW YORK
JOHN WILEY & SONS, Inc.
Lonpon: CHAPMAN & HALL, Limirep
1929
Copyricnat, 1929,
BY
Keira Henney
All Rights Reserved
This book or any part thereof must not
be reproduced in any form without
the written permission of the publisher.
Printed in U. 8. A,
PRESS OF
5/31 BRAUNWORTH & Co., INC.
BOOK MANUFACTURERS
BROOKLYN, NEW YORK
PREFACE TO THE FIRST EDITION
Third Printing Revised
Tus textbook on radio has been written for those who must
study without a teacher as well as for those who attend schools
where courses in radio are given. Every attempt has been made
to illustrate it with problems and examples which are practical in
nature; they deal with the values of electrical constants which the
radio engineer encounters. The experiments have been planned to
give the “‘feel’”’ of the apparatus that the research engineer or
_ experimenter uses.
The radio art moves forward rapidly and at times unexpectedly.
In this printing will be found material on tubes which have made
their appearance since the book was first published, a new and
simpler discussion of detection, and some description of technical
aspects, discussed in the first edition as only then in sight, are
handled in a more thorough manner because they are now in
general use.
The author desires to acknowledge his appreciation for the
aid which Howard E. Rhodes has given him in gathering material
for the text and for the aid which Robert S. Kruse rendered by
reading and criticising the original manuscript. Photographs,
drawings, and other material are included by permission of the
General Radio Company, the Samson Mfg. Co., E. T. Cunning-
ham, Inc., Proceedings of the Institute of Radio Engineers, and
Proceedings of the Institute of Electrical Engineers.
March, 1931. THE AUTHOR.
ee - a
a.
a ale
. eee
(tee 3 oe a ine en
ae a
CHAPTER
I
Il.
Ill.
Ve
CONTENTS
FUNDAMENTALS
The electron—Charged bodies—The laws of electrical charges
—The atom—The ether—The electric current—Insulators and
conductors—Conductivity—Resistance—The ohm—The effect
of molecular motion on resistance—The effect of temperature on
resistance—Temperature coefficient of resistance—The ampere—
The volt—Engineer’s shorthand—Mathematics in the study of
radio—Curve plotting—Symbols.
Oxum’s Law
Ohm’s law—Ways of stating Ohm’s law—Voltage drops—
Graphs of Ohm’s law—Series and parallel circuits—Character-
istics of parallel circuits—More complicated circuits—Detection
and measurement of current—Ammeters—Voltmeters—Sensi-
tivity of meters—Ammeter-voltmeter method of measuring resist-
ance—Voltmeter method of measuring resistance—Use of low
resistance voltmeter and milliammeter in high resistance cir-
cuits—Resistance measurement.
PRODUCRIONT OM CURRENT an aon yee ieee eer eerie
Batteries—Electrolysis—Common dry cell—Storage cell—
Internal resistance—Polarization—Cells in series—Cells in par-
allel—Magnetism—Oersted’s experiment—Faraday’s discovery
—tThe electric generator—Work done by alternating current—
D.-c. generator—Internal resistance—Electrical power—Power
lost in resistance—Efficiency.
TEN DUCTIA'N CWicusysteg: ee cote utero eintccd: « eae) stator eet obestaciene, > arctan sneaks
Coupled circuits—Lenz’s law—Inertia—Inductance—Self-
inductance—Magnitude of inductance and induced voltage—
The unit of inductance—Typical inductances—Coupling—
Magnitude of mutual inductance—Measurement of inductance
—The transformer—Power in transformer circuits—Transformer
losses—The auto-transformer—Transformer with open-circuited
vil
20
38
57
Viil
CHAPTER
Wile
WANE,
VIII.
CONTENTS
PAGE
secondary—Variable inductances—LEffect of current, frequency,
etc., on inductance.
CORD ACUEY -va lds F te cies OE aoe ore rae repairers oie ra
Capacity—Capacity as a reservoir—Capacity in a power sup-
ply device—The charge in a condenser—The quantity of elec-
tricity in a condenser—Energy in a condenser—Blectrostatic
field—Condensers in a.-c. circuits—Power loss in condensers—
Condenser tests—Condensers in general—The nature of the
dielectric—Sizes of radio condensers—Condenser capacity for-
mulas—Condensers in series and parallel.
PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
Definitions used in a.-c. circuits—Instantaneous value of alter-
nating current—Triangle functions—Means of expressing in-
stantaneous values—Effective value of alternating voltage or
current—Phase relation between current and voltage—Current
and voltage in phase—Lagging current—Inductive reactance—
Leading current—Capacity reactance—Comparison of inductive
and capacitive reactance—-Impedance—General expression for
impedance—Series a.-c. circuits—Phase in series circuit—Char-
acteristics of a series circuit—Resonance—Parallel circuits—
Phase in parallel circuits—Impedance of a parallel circuit—Series
and parallel circuits compared—Power in a.-c. circuits.
RESONANCE
Series resonant circuit—Characteristic of series resonant, cir-
cuit—Effect of resistance on series resonant circuit—Power into
resonant circuit—The resonant frequency of the circuit—Wave-
length—Parallel resonance—Effective resistance—Resonant fre-
quency—Uses of series and parallel resonant circuits—Sharpness
of resonance—Selectivity—Width of resonance curve—Fffect of
inductance and capacity on sharpness of resonance—The resist-
ance of coils—High-frequency resistance—Distributed capacity
of coils.
PROPERTIES OF Corts AND CONDENSERS
Tuning a receiver—The wavemeter—Heterodyne wavemeter
—Calibrating a wavemeter—Standard frequencies—Calibrating
by ‘‘clicks’”—Coil and condenser properties—Measurement of
coil resistance—Determining capacity of a condenser—To
89
120
149
CHAPTER
Ix.
Dole
CONTENTS
1X
PAGE
measure wavelength of an antenna—To measure capacity of
an antenna—Measure antenna inductance—A typical receiving
circuit.
PSEA C UniMon Ur El eer eee in cee SO) Carel ie MO OE satiny kp
The construction of the vacuum tube—The purpose of the
filament—The purpose of the plate—Effect of filament voltage—
Saturation current—The purpose of the grid—Characteristic
curves—Grid voltage—Plate current curves—Plate voltage—
Amphfication factor—The meaning of the amplification factor—
Equivalent tube circuit—D.-c. resistance of the tube—Internal
resistance of tube—The mutual conductance of a tube—Impor-
tance of mutual conductance—Slopes as tube constants—The
“Tumped”’ voltage on a tube—Measurements of vacuum tube
constants—Bridge methods of determining tube factors—To
measure the plate resistance—An a.-c. tube tester—Types of
tube filaments—Reactivating thoriated filaments—Alternating-
current tubes—Heater types of tubes—Operation of a.-c. tubes—
Operating filaments in series—Compensating for plate current
flowing through filaments—Means of obtaining C bias in amplifier
tubes—Sereen grid tube—Characteristic curves of the screen-
grid tube—Space charge grid—Mistreatment of tubes.
FEE RGOB ERAS @AIN 2 A MPLUEBIEIR stscsei sc) tuscceeien aie ero ero ae
The tube as an amplifier—Resistance output load—Dynamic
characteristic curves—Phase of H, H, I,—Magnitude of the
amplified voltage—Equivalent tube circuit—Power output—
Power amplification—Power output proportional to input grid
voltage squared—Amplifier overloading—Distortion due to
curved characteristic—Permissible grid swing—Distortion due
to positive grid—Amount of distortion caused by overloading—
Power output calculation—Harmonic distortion calculation—
Power diagram.
ACDIOSARMPD LETH RS tener etre ret entries cael ya tener siete nterstenercreisiecslre at:
Need of an audio amplifier—The requirements of an audio
amplifier—Cascade amplifiers—Effect of leaky condenser—Fre-
quency characteristic of resistance amplifier—Overall amplifica-
tion—Plate battery requirements—Inductance load amplifier—
Effect of stray capacities at high frequencies—Quantitative effect
of capacities on high frequencies—High-frequency response in
impedance-coupled amplifier—Tuned inductance amplifier—The
202
228
x
CHAPTER
ee
XIII.
XIV.
CONTENTS
PAGE
transformer-coupled amplifier—Transformer with no secondary
load—The advantage of the transformer—Reflex amplifiers—
The inverse duplex—Transformer-coupled amplifiers—Measure-
ments on transformer-coupled amplifiers—Calculation of overall
voltage amplification—‘ Equalizing”—The power amplifier—
The push-pull amplifier—General conditions for voltage and
power amplification—Power necessary for good loud speaker
operation—Uses of tubes in parallel.
Dersian or Aupio FREQUENCY AMPLIFIERS
The transmission unit—Voltage and current ratios—The use
of the DB—Design of audio-amplifiers—Transformer working
out of high impedance—Rules for the amplifier design—Com-
parisons between amplifiers—Volume control—Proper C bias for
tubes—Loss in output transformer—Manner of coupling tube to
load—Method of connecting choke condenser output—Voltage
limits on the output condenser—Regeneration in audio ampli-
fiers—A case of regeneration—Filtering in audio amplifiers—
Individual transformer characteristics—Comparison of push-pull
and single tube—Screen-grid audio amplifier.
HIGH-FREQUENCY AMPLIFIERS
Purpose of radio-frequency amplification—Field strength—
Advantage of high power at transmitting station—The task of
the radio-frequency amplifier—The ideal response curve of a
recelver—Three types of radio-frequency receiving systems—
Radio-frequency amplifiers in general—Effect of tube input
capacity—Tuned radio-frequency amplifiers—Effect of negative
input resistance—Engineering the tuned radio-frequency ampli-
fier—Gain due the tube and gain due the coil—Effect on second-
ary resistance of close coupling—Selectivity—Summary of radio-
frequency amplifier phenomena—Selectivity to signals far off
resonance—Uses of several stages of radio-frequency amplifica-
266
296
tion—Coil factors—Turns ratio into detector tube—Regen- °
eration and oscillation in radio-frequency amplifiers—Lossers—
Bridge systems—The neutrodyne—Neutralizing bridge circuits
—Filtering radio-frequency circuits—Use of screen-grid tubes
as radio-frequency amplifiers.
Drrrection
Distorting tubes—Modulation—Percentage modulation—De-
modulation—The plate circuit detector—Conditions for best
335
CHAPTER
EXCVA
XVI.
exeViUIE
CONTENTS
Xl
PAGE
detection—The vacuum tube voltmeter—Adjusting a volt-
meter—D.-c. plate current as a function of a.-c. grid voltage—
Detection in a radio-frequency amplifier—Grid leak and con-
denser detector—Effect of grid leak and condenser values—
Detector action—Power detection—Distortion from square law
detector. _
UE CHLVIN GAOYETEMS:75 0 -<trah cats oon ne a eeoai: a e n
The tuned radio-frequency set—The superheterodyne—The
phenomenon of beats—Superheterodyne design—Radiola 60
series—Repeat points—Choice of the intermediate frequency—
Selectivity of superheterodyne—Frequency changers—The auto-
dyne—‘‘Short-wave’’ receivers—Short-wave receiver circuits—
Coupling the short-wave receiver to the antenna—Use of screen-
grid tube at short wayves—Long-wave receivers—Detuning loss
in autodynes—Poor quality on long waves—‘“‘Band pass”’ re-
ceivers—The Sparton receiver—Experiments with band pass
filters—Measurements on radio receivers—Signal generator—
Modern receivers—Apparent and real selectivity—Importance
of shape of condenser plates—Automatic tuning—Automatic
volume control—Shielded receivers—Loud speakers—The horn
type—The moving coil speaker—Baflles for dynamic speakers—
The electrostatic loud speaker—The telephone receiver—Loud-
speaker measurements.
RECTIFIERS AND PowmR Supply APPARATUS................
The fundamental rectifier circuit—Kinds of rectifiers—Typi-
cal filament rectifiers—Requirements for rectifier tubes—Single-
wave rectifier—Gaseous rectifiers—Characteristics of gaseous
rectifiers—The Tungar rectifier—The sulphide rectifier—Filter
circuits for tube rectifiers of the filament type—Regulation—A
typical rectifier—Filter system—Hum output—The voltage
divider—C bias for tubes operated from alternating current—
Engineering the voltage divider—Voltage regulation.
OsCIREATORS a EL RANSMIUDTER Sm OCC at wtets ete stoe aisarerar terrae
Oscillating circuits—Undamped or continuous oscillations—
The amplifier as an oscillator—Conditions for oscillation—Maxi-
mum oscillatory plate current—Effect of coupling—Dynamic
characteristics—Conditions for oscillation—Efficiency of an
oscillator—Harmonics—Power output of an oscillator tube—
Maximum power output of oscillator—Obtaining grid bias by
356
390
415
xii CONTENTS
CHAPTER PAGE
means of resistance leak—Practical circuits—Hartley oscillator
—Shunt-feeding oscillators—Other oscillating circuits—Adjust-
ing the oscillator—Frequency stability—Master oscillator sys-
tems—Crystal control apparatus—Frequency doublers—Self-
rectified transmitters—Adjusting the plate load to the tube—
Plate current when oscillator is connected to antenna—Keying
a transmitter—Too close coupling to antenna—Methods of
connecting osci)lator to antenna—Feeding power through trans-
mission line—Modulation—Amount of power required for
modulation—Modulation at low power—Distortion at receiver
due to complete modulation.
VL aeANTENNAS E CLCAN SMISSION a iC eee ee a 453
Radiation resistance—The radiation field—Calculation of the
received current—Types of antennas—Directional antennas—
Inductance and capacity of antennas—Natural wavelength of
antenna—Loading an antenna—Decreasing the wavelength of
an antenna—Short-wave transmission—Fading—Comparison of
night and day reception—Static—Elimination of man-made
interference.
PRINCIPLES OF RADIO
CHAPTER I
FUNDAMENTALS
No one can learn a great deal about the theory and practice of
radio who does not also know a few fundamental facts about
electricity, for radio is but one aspect of a much broader field,
electrical engineering. And since electricity is but a movement
of the smallest known bit of matter and energy, the electron, it is
necessary that a study of electricity must be preceded by a slight
knowledge of the electron.
1. The electron.—The entire universe is made up of various
combinations of about ninety substances known as elements.
These elements are composed of but two things, negative electrical
charges known as electrons, and positive charges known as
protons. Theelectronsareallalike. The only difference between
copper and aluminum lies in the difference in the number and
position of their electrical charges.
2. Charged bodies.—The term charge is used in various ways.
A body on which there is an equal amount of negative and positive
electricity is said to be in equilibrium; but if the body has an
excess of either negative or positive electricity it is said to be
charged. Sometimes the body itself is called a charge and of
course may be referred to as a positive or a negative charge. If it
has a great excess of either of the two kinds of electricity, it is
said to be highly charged. In this condition it is in a state of
very unstable equilibrium, and at the least chance some change
will occur to bring the body into a state of greater equilibrium.
2 FUNDAMENTALS
3. The laws of electrical charges.—These electrical charges
obey simple laws: like charges, whether positive or negative, repel
each other; unlike charges, that is, a positive and a negative
charge, attract each other. The more highly charged the bodies
are the greater will be the repulsion or attraction. The closer
together the charges are the greater will be the attraction or repul-
sion. Doubling the distance between two unlike charges divides
their attraction by four. The greater the magnitude of the
individual charges the greater is the attraction or repulsion; the
greater the distance between the charges the less the attraction or
repulsion.
Experiment 1-1. This experiment demonstrates the production of charged
bodies by friction and the laws that control such charged bodies. Whenever
Fria. 1.—A familiar experiment in static electricity.
one body is rubbed by another, some frictional electricity is said to be gen-
erated. This amounts to stating that one of the bodies becomes charged. To
get an appreciable charge, one needs a glass rod, a stick of sealing wax, a piece
of silk and a piece of flannel, and a small bit of pith from a dry cornstalk or
alder branch or sunflower stalk suspended as shown in Fig. 1 by means of a
fine silk thread. Rub the piece of glass with the silk and bring near the pith
ball; the pith ball should be repelled with considerable force. Then rub the
wax stick with the flannel and bring it near the pith ball. It should be strongly
attracted, proving that another kind of static or frictional clectricity—or elec-
tricallycharged body—has been generated. Nowsuspend two similar pith balls,
touch one of them with the glass rod after rubbing it, and touch the other ball
with the wax rod. Bring them near each other. Touch both balls with either
the glass or the wax rod and bring them near each other. Now they are
THE ETHER 3
repelled. The latter part of the experiment is almost the starting point of the
world’s history of electrical experiment.
4. The atom.—The simplest form in which an element can
exist by itself is called the atom. A combination of two or more
atoms is called a molecule. Ordinarily the atom or molecule is in
electrical equilibrium with its surroundings. If, however, through
some severe mechanical shock for example, it should lose an elec-
tron it would be charged and then would follow the laws cited
above. It would then attract or get rid of an electron at the first
opportunity and become neutral again.
It is the motion of electrons that we know as the electric cur-
rent. When there is a sufficient number of electrons, a billion
billion per second, for instance, there is current enough to light an
incandescent lamp or heat an electric iron.
The atoms and molecules in matter are in constant motion,
carrying with them in their movements the electrons that con-
stitute them; in the bumping of one atom against another, elec-
trons are lost, gained, and interchanged.
Atoms of matter are inconceivably small. Everyone has seen
many-colored oil films on the street. It is possible to obtain oil
films less than half a ten-millionth of an inch thick. The atoms
composing these films cannot be thicker than this figure; the elec-
trons are much smaller yet. We think the distances in the solar
system of which the earth is part are beyond comprehension, the
sun for example being about 90 million miles from the earth; but
the dimensions of the electrons in their smallness are even more
difficult to picture. The diameter of the electron is estimated to
be about 1 foot divided by a hundred million million. Each of
these electrons resembles its brother exactly, so that when an elec-
tron is knocked out of an atom by a collision it is free to combine
with any other body near by which may have a deficit of negative
electricity, no matter what the body may be made of. The electron
is the unit out of which everything is made.
5. The ether.—The fact that one charge can exert a force,
either of attraction or repulsion, upon another implies that some-
thing connects the two. For instance, a comb which has been
rubbed on the coat sleeve will pick up bits of paper even though
4 FUNDAMENTALS
it does not actually touch them, the paper jumping to the comb
while the latter is still some distance from it. Evidently some-
thing exists in the space between the comb and the paper. That it
is not air may be demonstrated by performing a similar experiment
under a jar from which all the air has been pumped.
This leads us to a conception of what is commonly known as the
ether. It is simply the place or the substance, or whatever one may
choose to call it, wherein the attraction or repulsion of electrical
charges exists. The ether is an invention made necessary by our
difficulty in conceiving how one body can exert an effect upon
another except through some intervening medium. Between two
charged bodies are said to exist lines of force which tend to decrease
the distance between the bodies if they are oppositely charged or
to increase it if the bodies are charged alike. The sum of these
lines of force is called an electrical field and every charged body is
surrounded by such a field. Since a wireless aerial is but a charged
system of wires it too has a field about it. This field extends in all
directions through what we call the ether.
6. The electric current.—In an ordinary piece of copper wire
the electrons are moving about in a haphazard fashion at the rate
of about 35 miles per second. If this wire is in an electrical circuit,
in addition to this to and fro motion there is a comparatively slow
drift of electrons from one end of the wire to the other. It is this
slow drift of electrons in a given direction that we ordinarily call
the electric current. Because each electron can carry an extremely
small quantity of electricity, it is only movements of large numbers
of them that we are interested in. It has been estimated that it
would take all the inhabitants of the earth, counting night and
day at the highest rate of speed possible, 2 years to count the
number of electrons which pass through an ordinary electric light
ina second. This is about the same number that are necessary to
light the tubes in a four-tube battery-operated receiver.
The flow of electrons from one end of the wire circuit to another
can be explained by the two fundamental laws of electrical charges
(Section 3). When a wire is attached to the positive terminal of a
battery there is a momentary movement of the electrons nearest
the end of the wire toward the battery. This movement soon
INSULATORS AND CONDUCTORS 5
ceases because the flow of electrons into the battery leaves a dearth
vf them at the other end of the wire which must be supplied. If
both ends are attached to the battery a steady drift of electrons
_ takes place out of the negative pole or terminal of the battery,
through the wire, and to the battery again at the positive
terminal. 2
Thus the actual motion of the electrons is from the negative
toward the positive end of a circuit. This is in the opposite direc-
tion from the rule established many years before the electron had
been discovered, namely, that current flows from positive to nega-
tive. In problems the student can assume either direction so long
as he is consistent. We know that the electrons flow from negative
to positive; electrical workers assume the current flows from posi-
tive to negative. In this book we shall follow the latter rule, but
the student should remember that the actual carriers of electricity
move in the opposite direction.
7. Insulators and conductors.—It is a matter of common
knowledge that the current does not flow through the non-metallic
parts of a radio set, or through the insulating material around a
broken conductor. How does it happen that some materials are
such good ‘‘ conductors,”’ copper and silver for example, whereas
others, such as glass or bakelite, appear not to conduct at all?
Here again we are dealing with the building stones of all matter,
the atom and the electron. Atoms of the so-called non-conductors
maintain their hold on their individual electrons very tightly; few
electrons escape. In the conductors, the electrons of the various
atoms are freer to move about, and so to be interchanged among
atoms. A conductor is a substance that quickly loses its charge
when rubbed. A non-conductor retains its charge longer.
A good conductor is a material whose electrons are freer to
move about than those of a poor conductor. Strictly speaking,
there are no non-conductors. All materials will carry current to
some degree. Glass, for example, which is generally considered a
very good insulator, conducts electricity fairly well when it is in
a molten state.
The best insulators—that is, the poorest conductors—are
amber, rubber, sulphur, shellac, porcelain, quartz, silk, air. Dry
6 FUNDAMENTALS
wood, paper, cotton and linen thread are semi-conductors. The
best conductors are the metals, acids, moist earth, etc.
8. Conductivity—All materials have a certain characteristic
which we may call conductivity, which describes their ability to
conduct an electric current. Among pure metals silver has a
very high conductivity, copper is next, and near the bottom lies
iron with about one-ninth the conductivity of copper. The con-
ductance of a circuit is a term expressing its ability to pass an
electric current. The greater the conductance, the greater the
current.
9. Resistance.—Those metals which have a high conductivity
may be said to offer little resistance to the flow of electrons through
them. Thus, copper has
d “Heer a low resistance, while
te some combinations of
copper, nickel, and iron-
sae Las manganese, for example,
se + have resistances many
Fe its times that of copper. A
Lan device added to a circuit:
to increase its resistance
is called a “resistor.”
The words resistance and
f : 5 ; 4 resistor are used synon-
Fria. 2.—Resistance depends upon the length yinouely a air
and size of a conductor. The resistance of two
wires of the same material
and at the same temperature depends upon two things, the length
of the wires and the area of their cross-section. Naturally, the longer
the wire the fewer electrons can pass through it in a given time;
similarly the smaller the diameter of a wire the greater the resist-
ance, just as you can get more gallons of water per second from a
53-inch fire hose than from a l-inch garden hose, although they
may be attached to the same hydrant.
Similarly, a wire 2 feet long has twice the resistance of a wire
1 foot long but of the same diameter. Of two wires the same
tength the one having the smaller diameter will have the greater
resistance. (See Fig. 2.)
THE OHM 7
The resistances of several metals compared to silver are as
follows:
Silverson cose cee 1.00 Blaster 7.20
Copperrme se eee Lett German silver...... 14.20
INTRON. 5 5 ooo ooo 1.87 iardisteeli 3. oor 13,5)
INiekela evens seks = eee 4.67 Mercury ae een OStL
SOIL TRONS 6 ode kia aoee 6.00
Problem 1-1. How many times higher in resistance is mercury than
silver? Than copper?
Problem 2-1. Two wires of the same length and diameter have resist-
ances in the ratio of 5.45 to 1. If the lower resistance wire is copper, could
you identify the other wire material from the above table?
Problem 3-1. Two wires, one of soft iron and the other of hard steel, are
_to have the same resistance. They have the same diameter. The hard steel
wire is one foot long. What is the length of the soft iron wire?
10. The ohm.—The unit of resistance is the ohm. It is the
resistance of a column of mercury weighing 14.4521 grams, having
a uniform cross-section and a height of 106.3 cm. at 0° Centigrade.
A 9.35-foot length of No. 30 copper wire has a resistance of about
one ohm. The table on page 17 gives sizes and resistance per
thousand feet of copper wire. The resistance per foot may be
obtained from such a table by dividing the resistance per thousand
feet by one thousand.
Note that increasing the size of wire by three numbers, that is,
from No. 20 to No. 23, doubles the resistance of the wire from
10.15 to 20.36 ohms; going from No. 30 to No. 27 lowers the
resistance from 103.2 to 51.5 ohms per thousand feet.
Copper is used in electrical and radio circuits because of its
high conductivity compared to other metals and its low cost com-
pared to metals of higher conductivity. It is readily obtainable
and easily worked.
Problem 4-1. What size of soft iron wire will have the same resistance as
No. 32 copper?
Problem 5-1. What is the resistance of 1 foot of No. 20 copper? Of
No. 24 hard steel?
Problem 6-1. What is the resistance of a column of mercury of the fol-
lowing dimensions, weight 0.032 Ib. (1 pound = 453.6 grams), length 41.7
inches (1 inch = 2.54 centimeters) at 0° C., the column having uniform cross-
section?
8 FUNDAMENTALS
11. The effect of molecular motion on resistance.—Why do
some substances have greater resistance than others? Let us again
consider the electrons, atoms, and molecules which make up the
wires carrying the currents. Not only are the electrons in motion,
but the atoms and molecules themselves are in a sluggish motion,
the violence of this motion depending upon the temperature of the
wire and the material of which the wire is made.
And although molecules cannot traverse the electric circuit as
the electrons can, in their to and fro motion they impede the
progress of the electricity bearers by countless collisions with them.
The greater this molecular motion, the greater the resistance to a
progressive flow of electrons, and the greater the wire’s electrical
resistance.
12. The effect of temperature on resistance.—The resistance
of all pure metals rises with increase in temperature. This is be-
cause of the greater molecular agitation at higher temperatures,
making it more difficult for the electrons to drift in their progressive
motion around the circuit.
At absolute zero, 273 degrees below zero Centigrade, all molec-
ular motion is supposed to stop, making the resistances of metals
practically zero. At the lowest temperature reached it has been
found that the resistance of a coil of wire is so low that current
will flow for some time after the driving force is removed. Absolute
zero has not been reached up to the present time.
13. Temperature coefficient of resistance.—Conductors in
which radio engineers are interested increase or decrease in resist-
ance at a regular rate with respect to temperature. The change
in resistance of a given wire may be computed from the following
facts. The “temperature coefficient of resistance ”’ is a term
which gives the amount a resistance increases for each degree rise
in temperature for each ohm at the original temperature. For
example, if a copper wire with a temperature coefficient of 0.0042
has a resistance of 80 ohms at 0° C., this resistance will be increased
by 80 X 0.0042 for each degree rise in temperature. At 50° C. the
resistance increase would be 80 X 0.0042 X 50 or 16.8 ohms, and
the resistance would now be 80 + 16.8 or 96.8 ohms. Manganin
wire, composed of 84 per cent copper, 12 per cent manganese,
THE AMPERE 9
4 per cent nickel, has a very low temperature resistance. It is
0.000006. The change in resistance of two metals with temperature
is shown in Fig. 3.
14. The ampere.—The ampere is the term used to express the
rate at which electrons move past a given point in an electrical
circuit. It is equal to 6.28 10!® (see Section 16) electrons per
second. Since each electron carries a definite quantity of elec-
tricity, the total amount carried by 6.28 X 10!8 electrons is a
% Change in Res.
Nichrome = Nickel-Chromium Alloy
S.M.L.=67% Ni; 28% Cu; 5% Fe,Mn, Si, etc.
Temperature Degrees C.
Fic. 3.—Effect of temperature on resistance.
definite quantity and is known as the coulomb. The ampere, how-
ever, is the term used in electrical practice. It corresponds to the
term gallons per second used in speaking of the amount of water
transported through a pipe or hose. The term gallons alone con-
veys little meaning, since the same number of gallons will flow out
of a small hose as out of a large one provided we do not consider
the time involved, or the pressure. But “ gallons per second ”’ in-
volves both time and pressure and is a term easily visioned. A
current of one ampere will convey through a circuit one coulomb
of electricity per second.
10 FUNDAMENTALS
The ampere as a quantity of electricity transported per second
is a large unit if we compare it with the current flowing from the
B batteries of a radio set. It is small compared with the currents
encountered in power houses. The B batteries supply only
thousandths of amperes or milliamperes, whereas in a small power
house supplying power to a village one may have thousands of
amperes flowing. A meter to measure the flow of current is called
an ammeter, or milliammeter, or microammeter, depending upon
the strength of current it can measure. Approximate currents
flowing through commonly used devices are shown below.
Approximate
Apparatus Current in Amperes
SOSwWia titel ann eee ace Meee eee 0.5
2OU=Wavu LAM mrene aer anteater 2.0
2-horsepower motor............... 10
(leCtri Call Onae ae ener eee 5
Filament of vacuum tube.......... 0.25
Plate circuit of vacuum tube........ 0.001
15. The volt.—The electrons are driven through the wires and
apparatus composing the circuit by a force called an electromotive
force, abbreviated to e.m.f. The unit of force is known as the
volt. It is the electrical force that will cause one ampere of elec-
tricity to flow through a wire which has one ohm of resistance.
The common dry cell used to ring door bells has a voltage of about
1.5; storage batteries when charged have a voltage of about 2.0
and thus a three-cell battery has a voltage of 6.0. The ordinary
B battery has a voltage of about 45, and if torn apart will be found
to consist of 30 small cells, the voltage of which is 1.5 volts each.
When the total voltage of such a battery is as low as 37 the battery
should be thrown away. An instrument used to measure voltages
is known as a voltmeter. A table of voltages is given below.
Apparatus Voltage (Approximate)
Dry “cell 230 iee tere che eee ib 5)
Storage: batter yagi ae 6
B battery. sac nee eee 45
louse ig htine tei cul eee ener n ane 115
sorthird rail?’ Se eeniect tore eens 500
ENGINEERS’ SHORTHAND valu’
16. Engineers’ shorthand.—Engineers have a simple short-
hand method of working with large numbers well illustrated by
the figures 6.28 X 10!8, which really means that 6.28 multiplied
by a million million million electrons flowing past a given point
_ per second constitute the electric current known as an ampere.
We shall have occasion to use this system many times in the course
of the book and students are encouraged to master it as soon as
possible. The table below will be helpful.
1 = 10° = one
10 = 10! = ten
100 = 10? = hundred
1000 = 10? = thousand, ete.
1 = 10° = one
1 = 1071 = + = one-tenth
.01 = 10-2 = =45 = one-hundredth
.001 = 10-* = yao = one-thousandth, ete.
The small number above the figure 10 is called the exponent.
Numbers less than 1 have negative exponents. Thus three-
thousandths may be expressed in these several ways:
3 3
003 = 3 X10 = = 55
When numbers are multiplied, their exponents are added;
when the numbers are divided, the exponents are subtracted.
Thus 100 multiplied by four-tenths may be done in shorthand
as follows:
LOO S< 45 1025645 —K 1052
= 4 S< ie
=4x 10
= 40
Similarly, let us divide 3000 by 150.
3000 + 150 = (3 X 10%) + (1.5 X 102)
= ws «x 10? Kk 10
5
2x 10
= Ai)
ll
12 FUNDAMENTALS
The rules are few and those are simple:
1. To multiply, add exponents.
2. To divide, subtract exponents.
3. When any number crosses the line, change the sign of the
exponent.
Example 1-1. Multiply 20,000 by 1200 and divide the result by 6v00.
20,000 = 2 X 104
1200 = 12 X 10?
6000 = 6 X 103
20,000 X 1200 2 x 104 x 12 X 10?
6000 6 X 108
Ziel 2s a Ooe <a O29 alae
- 6
= 2gt x 103
= 4000
Problem 7-1. How many electrons flow past a given point per second
when the number of amperes is 6? 60? 600? 0.1? 0.003?
Problem 8-1. The sun is roughly 90 million miles from the earth.
Express this in ‘ shorthand.”
Problem 9-1. At 100 miles per hour, how many months would it take to
reach the sun?
Problem 10-1. [If light travels at 300 million meters per second and if a
meter equals 3.3 feet, how long does it take the sun’s rays to reach the earth?
Problem 11-1. How many amperes of current flow when 31.4 « 10%
electrons per second flow past a point?
In connection with such shorthand methods the following
table of prefixes commonly used will be important.
Prefix Abbreviation Meaning
micro mm one millionth
milli m one thousandth
centi c one hundredth
deci d one tenth
deka dk ten
hekto h one hundred
kilo k one thousand
mega M one million
CURVE PLOTTING 13
Thus a thousandth of an ampere is known as a milliampere, a
million ohms is called a megohm, etc.; or, expressed in numbers,
1 milliampere = 10-* or .001 ampere; 1 megohm = 1,000,000
ohms.
17. Mathematics in the study of radio.—The amount of mathe-
matics one needs varies with the intensity with which one intends
to study radio. As in all other branches of science in which
mathematics plays a part the easier it is to think mathematically
Y Axis or
Axis of Ordinates
X Axis or
Axis of Absissae
Fia. 4.—Curves are usually drawn with the origin as
shown.
the greater are the possibilities ahead of the student. In this book
it is necessary to have only a rudimentary knowledge of algebra to
work most of the problems. The student with no mathematics
beyond arithmetic and common sense will be able to work his way
through most of the examples.
18. Curve plotting —Many of the answers to radio problems
can be seen visually if the problem is plotted in the form of a
graph. Such graphs or curves are used frequently in this text, and
it is essential that the student and experimenter not only shall be
14 FUNDAMENTALS
familiar with how to plot curves but how to interpret curves that
other experimenters have
drawn.
The simplest form of
curve is a map. The map
has two co-ordinates or
axes, north-south and east-
west. Wesay thata town
is situated at so many miles
east and so many miles south
of some point that we take
as the origin. We have now
plotted the simplest mathe-
matical equation,a point. A
railroad runs straight north
past a point, that is,so many
miles east of this town. We
Fic. 5.—The origin in the center. can locate this railroad by
giving two points through
which it passes, or by giving one point through which it passes
and its direction. By a point and a direction we have plotted the
next simple mathematical
equation, a straight line.
In radio plots the axes
may be called X for the
horizontal and Y for the
vertical, or current for
one and voltage for the
other, etc. A graph isa
visual expression of the
relation existing between
two factors, X and Y, or
current and voltage, etc.
When one increases the
—
other increases or de- Fia.6.—How tocalculate the shape of a curve.
creases. Knowing the
law connecting them (the equation or formula, we call it) we can
CURVE PLOTTING 15
tell what the current is at any given voltage. If the law expressed
visually in the form of a graph is a straight line, we say the two
factors, current and voltage, are proportional. If one increases
and the other decreases, we say they are inversely proportional.
D. C. Output Voltage at ‘‘v’’
J0090009
115 V. 60.
125} |
20 40 80 100 120
60
D. C. Output MA.
Fria. 7.—A “B eliminator” regulation curve.
Curves are useful not only in giving us a visual picture of what
is happening in a circuit, but in telling us if the figures secured in
an experiment are correct. Thus we may calibrate a wavemeter
in condenser dial degrees against wavelength in meters. We plot
this curve and one or more points do not seem to fall on the smooth
16 FUNDAMENTALS
curve that goes through the other points. Something in our
laboratory experiment caused these points to be off the curve.
They were incorrectly taken, and the measurement that gave us
these points must be repeated.
The origin, sometimes, is at the lower left-hand corner of the
curve, as in Fig. 4, although it may be in the center or somewhere
else, as in Fig. 5, which shows the relation between the current in
the plate circuit of a vacuum tube as it is controlled by the amount
of voltage on the grid of that tube.
Points which lie to the left of the vertical axis are negative;
D. C. Output Voltage
MA Output
Fig. 8.—The same data as in Fig. 7 but to a different
scale.—Note how much flatter the curves appear.
points which lie below the horizontal axis are negative. All others
are positive.
The change in the vertical units with a given change in hori-
zontal units is called the slope of the curve. If the curve goes
through the origin this slope amounts to the ratio between the
vertical and the horizontal values at any point. In Fig. 6 is
shown the method of calculating the slope.
The units in which a curve is plotted change its appearance.
Thus in Fig. 7 is plotted the relation between the output voltage
of a “B eliminator” as the current taken by the receiver is
changed. Figure 8 shows the same data but plotted to a different
vertical scale. The slope of these lines looks different but really is
the same. If the slope is the factor in which we are interested, the
SYMBOLS 17
more open scale should be used so that small changes will be
visible.
19. Symbols.—In all technical literature a number of abbrevi-
ations are used to express parts of circuits. For example, in very
popular articles a “‘ picture diagram ”’ may be used, but picture
diagrams are only for the boy and the experimenter who are too
lazy or disinterested to learn radio language. The symbols used in
this book are shown on the next two pages. A circuit is built up
simply by connecting several of these symbols together.
Coprer WIRE TABLES
Resistance at 68° F. (20° C.)
Mils, .001 inch
Size of ;
Wire | Diameter | Ohms per |Pounds per pene eer e oe
B. & S| of Wire 1000 Ft. 1000 Ft.
Gauge Mils. S.c.c. D.c.c. S.s.c. D.s.c.
0000 460 -049 | 641 2.14 2.10
000 410 -0618 | 508 2.39
00 365 -0779 | 403 2.68 2.62
0 325 -0983 | 319 3.00
289 .1239 | 253 3.33 3.25
2 258 -1563 | 201 3.75
3 229 -1970 | 159 4.18 4.03
4 204 . 2485 | 126 4.67
5 182 .3133 | 100 Soe 5.00
6 162 .3951 79.5 5.88
vs 144 -4982 63 6.54 6.25
8 128 - 6282 50.0 7.35
9 114 7921 39.6 8.26 7.87
10 102 .9989 31.4 9.25
11 91 1.260 24.9 10.3 9.80
12 81 1.588 19.8 11.5
13 72 2.003 15.7 12.8 12.2
14 64 2.525 12.4 14.3
15 57 3.184 9.86 15.9 14.9
16 51 4.016 7.82 Ti. 16.7 18.9 18.3
17 45 5.064 6.20 20.0
18 40 6.385 4.92 22.2 20.4 23.6 22.7
19 36 8.051 3.90 24.4
20 32 10.15 3.09 27.0 24.4 29.4 28.0
21 28.5 12.80 2.45 29.9
22 25.3 16.14 1.94 33.9 30.0 36.6 34.4
23 22.6 20.36 1.54 37.6
24 20.1 25.67 1.22 41.5 35.6 45.3 41.8
25 17.9 32.37 Ae 45.7
26 15.9 40.81 . 769 50.2 41.8 55.9 50.8
27 14.2 51.47 .610 55.0
28 12.6 64.90 .484 60.2 48.6 68.5 61.0
29 11.3 81.83 . 384 65.4
30 10.0 103.2 . 304 71.4 55.6 83.3 72.5
3 8.9 130.1 241 (ihe)
32 8.0 164.1 191 ae 62.9 101 84.8
33 Tiel 206.9 152 90.
34 6.3 260.9 120 97.1 70.0 121 99.0
5.6 329.0 0954 104
3 5.0 414.8 0757 111 (tif) 143 114
37 4.5 523.1 0600 118
38 4.0 659 .6 0476 ae 83.3 167 128
3s 3.5 831.8 0377 35
40 3.1 1049 0299 141 90.9 196 145
18 SYMBOLS
Galvanometer
Headphones
‘ i i k
Wires Connected Reversing Switch Closed Circuit Jac
| nels Y cae L= Inductance
C = Capacity
Single Pole
Wires crossed Double Throw Switch
but not Connected “S.P.D.T.” ®= Ohms
R= Resistance
—-O. = Megohms
mH= Millihenries
Mfd.= Microfarads
o————_ |[Mmfd.= Micro Microfarads
/H = Microhenries
Ammeter or “S.P.S.T.”” Switch
Milliammeter
——) Oo-—
——_—o o——
Voltmeter Double Pole
Double Throw Switch
74 Bt rl ® Yat Peel
SYMBOLS 19
tron Core
Transformer
Variable Resistance
Variable inductance Fixed Capacity
or Condenser
Variable Capacity
Variometer or Condenser Heater Type A.C. Tube
Transformer
with Antenna
Fixed Coupling
Four Element Tube
(Screen Grid Type)
M
Transformer
with
Variable
Coupling
Two Element or Rectifier
Tube
a
—.09090000-|—vwwww—=
Resistance or Impedance
Fixed Microphone
Iron Core
“Choke”
CHAPTER II
OHM’S LAW
In the previous chapter we stated that an electric current was a
motion of electrons; that the force which caused the motion of the
electrons was called an electromotive force (e.m.f.) or a potential
difference (p.d.), that the resistance of a circuit opposes the flow
of current, and that the unit of the current which actually flows
per second is called the ampere. We have then a very simple law
which enables the engineer to calculate
1. The current that will flow when the voltage and resistance
are known.
2. The voltage necessary to force a certain amount of current
through a known resistance.
3. The resistance that will restrict the current to a certain
value under pressure of a certain e.m.f. expressed in volts.
20. Ohim’s law.—The law which governs all simple and many
complex electrical phenomena is known as Ohm’s law. This law
states that: Current in amperes equals e.m.f. in volts divided by
resistance in ohms, or, as expressed in electrical abbreviations,
E (voltage)
I eS :
Cores, R (resistance)
21. Ways of Stating Ohm’s law.—There are three ways of
stating this fundamental law. They are:
(1) f= E/R (2) H=IXR (3) R= E/I
These three ways of stating the same law are determined from
the first statement of Ohm’s law by simple mathematical trans-
formation, and make less difficult the solving of problems.
20
VOLTAGE DROP 21
22. Voltage drop.—The second way of stating Ohm’s law is
that whenever a current flows through a resistance, there is a differ-
ence of potential at the two ends of that resistance. For every
ampere of current that flows through an ohm of resistance, there is
a volt lost. In other words it requires a volt to force an ampere
through an ohm of resistance, and once that task is over, the volt
is gone.
Consider Fig. 9, which shows the voltage-divider between a
radio receiver and a voltage supply system for feeding power to
the receiver. The power tube may require 180 volts. Other tubes
require only 90 volts or perhaps
small negative voltages. If 20 mil-
liamperes of current flow through
this voltage-divider, whose resist-
ance may be 5000 ohms, and To Rectifier
if there are 180 volts across the and Filter
entire resistance, there will be
other voltages along the resist-
ance as indicated in the illus-
+
tration. If the negative terminal ere
of a voltmeter were attached to p.. 9 The Poliee Ciiteons
the negative end of the voltage- receiver's power apparatus.
divider and the positive terminal ‘
of the meter touched to various points along the resistance on
the way toward the positive end, greater and greater voltages
would be measured. What is being measured at each point is the
drop in voltage between that point and the negative terminal of
the voltage-divider.
Often in laboratories a voltage is needed so small that it can-
not be measured with available instruments. A larger voltage can
be measured easily, however; and if it is impressed across a
voltage-divider, any desired part of the total voltage may be
utilized by tapping into this divider. (See Problem 4-2.)
These voltages appearing across a resistance because of current
flowing through that resistance are known as Ik drops. They may
be calculated by multiplying the resistance in ohms by the current
in amperes.
22 OHM’S LAW
Example 1-2. In Fig. 10 is an “ output choke ”’ coupling a loud speaker
to a power tube. The tube requires that 180 volts shall be impressed between
its plate terminal and its negative filament terminal. The question is, how
many volts must the B battery have in order to impress this voltage on the
tube?
The plate current is 18 milliamperes. The d.-c. resistance of the choke
coil is 1700 ohms. The voltage used up in forcing 18 milliamperes through
the 1700 ohms is 30.6. This voltage never gets to the tube. The voltage
measured between plate and negative filament will be the B battery voltage
minus the JR drop across the choke. Hence the total voltage required is 180
for the tube and 30.6 for the choke, or 210.6 volts. One of the requirements
of a good choke for such purposes is that it have a low d.-c. resistance. Other-
wise too much voltage is lost there.
i
I=.018 £, -180+.018x1700
=210.6
Fig. 10.—A_ choke-con- Fia. 11.—Use of an JR drop as a
- denser coupling circuit. source of voltage.
Problem 1-2. What current flows through a vacuum tube filament con-
nected to a 5-volt battery if its resistance is 20 ohms?
Problem 2-2. Suppose we desire to limit to 1 milliampere the flow of
current in a circuit attached to a 45-volt B battery. What must be the total
resistance of the circuit?
Problem 3-2. How much voltage is required to force a current of 60 milli-
amperes through a tube whose filament resistance is 50 ohms?
Problem 4-2. Consider Fig. 11. How many milliamperes of current must
be forced through the circuit in order to get 20 millivolts across the resistor
A-B? How many volts in all will be needed?
23. Graphs of Ohm’s law.—An interesting study of Ohm’s
law may be made by means of the circuit shown in Fig. 12 and
several sheets of plotting or graph paper. The result of plotting
current against voltage with constant resistance (Fig. 13); or
GRAPHS OF OHM’S LAW 23
resistance against current with constant voltage (Fig. 14); or
Fig. 12—A circuit for
testing Ohm’s law.
=
eB HO O FP OH DN BW
2 eS BB
E
Fic. 13.—In an Ohm’s law circuit
plotting current against voltage
results in a straight line.
voltage against resistance with constant current—all give an accu-
he
So
Ll
fat ND 00 te OT Oy) S100) 6
ae eo era e+ 1
Fie. 14.—The result of plotting current
against resistance.
rate graphical picture of what
Ohm’s law means. In Chapter
I the term conductance, K,
was defined. It is equal to 1/R.
When conductance against
current is plotted, a straight
line results, as shown in Fig.
15. When voltage and current
are plotted with a fixed resist-
ance in the circuit, the curve
is a straight line if the circuit
follows Ohm’s law.
Experiment 1-2. If the appa-
ratus is at hand, connect it up as in
Fig. 12, using a 6-volt battery, a
40-ohm rheostat, and an ammeter
reading a maximum of about 0.5 ampere. Connect the maximum resist-
ance in the circuit and note the current as the voltage is changed from
2 to 4 and then 6 volts by tapping onto each of the three cells of the
storage battery. Plot the data. Then use a smaller value of resistance
and repeat.
24 OHM’S LAW
Then use 2 volts and adjust the rheostat until several current readings
have been obtained. Calculate from Ohm’s law what the resistance is and
plot resistance against current.
10.0
ies A
. “
8.0
7.0
6.0
5.0
4.0
Current (1) in Amperes
3.0
2.0
1.0
ofl § af 73 4 43) -6 aif 8 Ae) 1.0
Conductance (K) in Mhos
1 F
Fig. 15.—Conductance (4) plotted against current.
Convert the resistances into conductances and plot against current. Cal-
culate similar data, using 4 volts and then 6 volts, and plot the data.
Note in each case the shape of the
curve has not changed, although the slopes
1=.02 of the straight lines vary with the resist-
ance and the curved lines are displaced
from each other.
Problem 5-2. Plate current (20 ma.)
flows through R in Fig. 16. What must
be the resistance in order to make the grid
ay aa of the tube 40 volts more negative than
Fre. 16.—A problem in Ohm’s the cathode? There is no loss in voltage
law—to determine the value of ™ the coil.
R to provide bias for grid of
tube.
24. Series and parallel circuits.—
There are two ways in which elec-
trical apparatus may be connected together.
SERIES AND PARALLEL CIRCUITS 25
When two or more pieces of equipment are connected as in
Fig. 17 they are said to be in series. The same current flows through
each unit. The voltage drop across each unit is controlled by its
resistance, and if one of these units has twice the resistance of the
other, the voltage drop across it will be twice as great. The sum
of the voltage drops across
the three resistances must
be equal to the voltage of
the battery, for there is no
other place in the circuit for
the voltage to be used.
In a series circuit the
total resistance is the sum
of the individual resistances. The current in each unit is the same
as in all other units. The current is obtained from Ohm’s law (1).
If any of the units becomes ‘‘ open ”’ the current ceases to flow.
If, however, any unit becomes “ shorted ”’ the current will increase
because the total resistance of the circuit has decreased.
Fic. 17.—A simple series circuit.
Example 2-2. In Fig. 18 is a typical series circuit composed of a vacuum
tube of the 201-A type which has a filament resistance of 20 ohms, a 6-volt
20 Ohms
Circular
Rheostat
Storage
Battery
Fig. 18.—A series circuit Fig. 19.—The equivalent
used in radio apparatus. of Fig. 18.
battery, a current meter, and a rheostat or variable resistor whose purpose is
to limit the flow of current through the filament of the tube. Note also
Fig. 19 in which the same circuit is represented using electrical symbols. The
arrow through R, indicates that it can be adjusted in value. fz» indicates
the resistance of the filament.
26 OHM’S LAW
The question is, what current will flow through the circuit as the resistance
of R, is varied? Suppose it is 4 ohms. We know the same current will flow
through both the filament and the rheostat. The resistance, then, in the
circuit is equal to 20 plus 4 or 24 ohms, and by Ohm’s law we know that the
current will be the voltage divided by the total resistance, or
There are two resistances in this circuit. Current flows through them.
There must then be two voltage drops. Let us calculate what they are. By
equation 2 we multiply the resistance by the current.
voltage drop = 7R, = .25 ampere X 4ohms = 1 volt
voltage drop = IR, = .25 ampere X 20 ohms = 5 volts
In other words, of the six volts available at the terminals of the battery,
five have been used up across the 20-ohm resistance and one volt has been
used to drive 0.25 ampere through the 4-ohm resistance.
Problem 6-2. Receiving tubes are often connected in series. Suppose
you have a 6-volt battery. How many 12-type tubes could you use in series
if each requires 1.5 volts across its filament?
Problem 7-2. Suppose you were going to use five 199-type tubes in a
series filament circuit. How many volts will be necessary?
Problem 8-2. What would be the total resistance of five 199-type tubes
in series?
Problem 9-2. The resistance of 199-type tube
filaments is about 50 ohms. How much current
would flow in Problem 8? Assume # = 3.
Problem 10-2. How much vesistance would be
necessary if one 199-type tube is to be run from a
6-volt storage battery? Two tubes? Five tubes?
Problem 11-2. An incandescent lamp has a
resistance, when hot, of about 55 ohms, and re-
E l= 001Ampers quires serrate: to light at full brilhaney. How
ROME —A problem many cou e run in series on a 110-volt cir-
V=90 Volts
R=100,000 Ohms
; : cult?
Pe ci, ee Problem 12-2. How many volts are required
er Sear at Is the +> force one milliampere through a circuit composed
value of Hi?
of a vacuum tube and a resistance, if the latter
has 100,000 ohms and 90 volts are required at the tube? (See Fig. 20.)
CHARACTERISTICS OF PARALLEL CIRCUITS 27
25. Characteristics of parallel circuits—A parallel circuit is
represented in Fig. 21. It consists of several branches. The
voltage across each branch is the same as
that across every other branch and is
equal to the voltage of the battery. The
total current supplied by the battery is
the sum of the currents taken by the
branches. ‘The resistance of the group
may be found by
R Ri R2 Rs Fic. Garis cir-
where F is the resultant or total resist-
ance, and Rj, Re, etc., are the individual resistances.
The resultant resistance of several units in parallel is less than
the individual resistance of any of the components. If two equal
resistances are in parallel, the resultant is one-half the resistance
of one. Thus if two 10-ohm resistances are connected in parallel,
the resultant resistance is 5 ohms. What would it be if they were
connected in series?
If any number of equal resistances are in parallel, the resultant
resistance is the individual resistance divided by the number of
units.
If only two unequal resistances are in parallel the resultant may
be calculated by dividing their product by their sum:
_ ki X Re
Ri + Re
Example 3-2. What is the parallel resistance of two units which have
resistances of 4 and 5 ohms?
This can be solved by either of the formulas given above.
|
ne
on
= 1/.45 = 2.22 ohms.
Sy
28 OHM’S LAW
_ RX Re
Ri + Rs
4X
Dad 45
i)
“a
9
oO
= 2.22 ohms.
Example 4-2. Suppose, as in Fig. 22, these two resistances in parallel are
placed in series with a resistance of 1 ohm and across a battery of 6 volts.
What current would flow out of the battery and through each resistance?
The total resistance is 2.22 + 1 = 3.22 ohms. The current flowing, then,
is 6 + 3.22 = 1.86 amperes. This current through the combined resistance of
—I,
R,=4
200 Ohms
R,
E=15 — 500 Ohms 800 Ohms
Fra. 22.—Solve for the various Fie. 23.—What is the voltage drop
currents. across the 800 ohms?
the 4- and 5-ohm units produces a voltage drop of J x R or 1.86 X 2.22 or
4.14 volts. This voltage across 4 ohms produces a current of 4.14 + 4 or
1.035 amperes, and across 5 ohms produces a current of 0.827 ampere. These
two currents added together are 1.862 amperes, which checks our calculation
above.
Problem 13-2. The usual battery-operated radio receiver has five tubes
of the 201-A type, each taking .25 ampere. What is their combined resistance,
and how much current do they take from a 6-volt storage battery if an external
rheostat is used to cut down the voltage to 5 across the tube? What is the
value of this resistance in ohms?
Problem 14-2. A circuit has three branches of 4, 6, and 8 ohms. A cur-
rent of 4 amperes flows through the 6-ohm branch. What current flows
through the other branches?
Problem 15-2. Consider the circuit of Fig. 23. What is the voltage drop
across the 800-ohm resistor?
DETECTION AND MEASUREMENT OF CURRENT 29
Problem 16-2. In Problem 13 suppose one of the five tubes were a power
tube requiring one-half ampere. If it is connected in parallel with the other
tubes, what will be the value of resistance needed? Will each tube get its
rated voltage under these conditions?
Experiment 2-2. Connect as in Fig. 22 several shunt resistances such as
tubes, rheostats, fixed filament resistors, etc., in series with a battery and a
rheostat. Measure the parallel resistance and individual resistances by read-
ing the current through them separately and in parallel and the voltage of the
1 1 1 1
battery. Test the relation — = :
Pr, eR.
3
26. More complicated circuits.—Some radio circuits are com-
binations of series and parallel circuits. A common form and its
equivalent are shown in Fig. 24. Other more complicated circuits
<——E >
Thy,
R,
ae E(R,+R3)
TOR, Ro+R,R3+R2R3
es ER,
Ry RL+R, Rs +RaR,
= ER,
2 R,R,+R,R, +RaR;
Fig. 24.—A complex circuit and its solution.
I,
I
may arise in practice and may be solved by more complex algebra
than is needed for simple Ohm’s law cases. All such circuits can
be reduced to more simple circuits by the application of certain
rules which may be found in books on complicated networks of
resistances, voltages, and current. In “Transmission Circuits for
Telephone Communication,” by K. 8. Johnson, may be found the
equivalent circuits of many very complex arrangements of
apparatus.
27. Detection and measurement of current.—We cannot see
or hear or smell the passage of an electric current through a cir-
cuit. It must be made evident to us by its effect upon the circuit.
There are three kinds, a magnetic effect, a chemical effect, and a
heating effect. Wire gets hot if too much current flows through it;
two dissimilar metals (copper and zinc, for example) placed in a
solution of one of them (copper sulphate) give off gas bubbles when
a wire connects them together externally; a wire carrying an
30 OHM’S LAW
electric current if brought near a compass needle will cause the
needle to change from its habitual north-south position.
These are the three fundamental effects of electricity. Any of
them can be used to detect the presence of a current or even to
measure the rate at which the current flows. A hot wire ammeter
(Fig. 25) for example is merely a wire which sags when it gets hot
by passing a current through it. A needle is attached to the wire
and moves across a scale as the wire gets hot. We might measure
the quantity of gas given off per unit of time and thereby deduce
the amount of current flowing
| through an electric “ cell.”
Most measuring instruments
| use the magnetic principle. They
| consist of a permanent mag-
; net near which is a coil of fine
| wire wound on a _ movable
pointer. Current flowing through
| this coil makes a magnet of
| it. It changes its position with
| respect to the permanent mag-
Fig. 25.—Hot wire ammeter. net just as a compass needle
does when brought near a cur-
rent carrying wire. Such instruments can be made sensi-
tive enough to measure currents as low as one-millionth of an
ampere or to detect the flow of even smaller currents than this.
28. Ammeters.—Meters to measure current are called am-
meters. They are connected in series with the source of current
and the device into which the current flows. They are made less
sensitive—so they will measure large currents—by shunting
them by copper wires so that only a small part of the total current
flowing actually goes through the meter.
A very simple current-indicating device consists of a coil of
wire through which the current flows and a compass placed in
the center. A modern highly sensitive meter is a delicate instru-
ment in which the compass needle is replaced by a carefully
pivoted coil of wire carrying a pointer. These instruments are
shown in Fig. 26.
VOLTMETERS dl
Fia. 26(a).—A simple form of galvanometer. When a current flows through
the coil of about 25 turns, a magnetic field is created which effects the position
of the needle.
ee
Fic. 26(6)—A modern meter (West-
inghouse type PX) which reads full
scale 200 microamperes and which will
indicate a current of less than 2 micro-
amperes.
29. Voltmeters.—Ammeters
have low resistance. They are
in series with the apparatus
taking current from _ the
source, as shown in Fig. 27.
Fic. 27.—Ammeters are connected
in series with the resistance into
which the current flows
Voltmeters, on the other hand,
must read the voltage across
some part of the circuit. They
must not permit much cur-
rent to flow because this cur-
rent would be taken away
32 OHM’S LAW
from the circuit. They have a high resistance. They are really
high resistance ammeters. Thus an ammeter, the Jewell 0-1 milli-
ampere meter for example, can be made to read volts by putting it
in series with a high resistance and across the circuit to be
measured.
For example, 1 volt is required to give 1 milliampere of
current through 1000 ohms. Thus if we have a 1.0 milliampere
meter and we wish to measure a voltage of the order of 1 volt,
we need only a resistance of 1000 ohms. Then the figures on the
meter scale will read volts instead of milliamperes. Such a series
resistance is called a multiplier.
A Weston Model 301 meter reading 1.0 milliampere full scale
will measure a maximum current of 50 ma. if a resistance of 0.57
ohm is placed across it. If the resistance is reduced to 0.27 ohm,
the meter will read 100 milliamperes when its needle points to
1.0 milliampere. Weston Models 280 and 301 voltmeters have a
resistance of about 62 ohms per volt. Thus a meter reading a
maximum of 100 volts has a resistance of 6200 ohms.
Problem 17-2. How much current is required for full scale deflection on
a Weston Model 301, 50-volt voltmeter?
30. Sensitivity of meters.—A sensitive current-measuring
meter is one which will measure very small currents but which has
a low resistance. A sensitive voltmeter is one which will give a
large needle deflection through a very high resistance. Voltmeters
which are used to measure the voltage of high resistance devices
such as B eliminators must have high resistance in order that the
current taken from the device shall not
be great enough to lower appreciably
E=100 x the voltage of the device.
IE cgemaatienres fT Sess
Fig. 28.—A low resistance
voltmeter placed at X will not
read the open circuit voltage.
R=10,000
Example 5-2. Suppose we are to measure
the voltage across the circuit at the point X in
Fig. 28. The voltage at X depends upon the
current taken by the meter. What is desired
is the “open-circuit ”’ or ‘“ no-load” voltage
across X, that is, the voltage existing there if no current is taken by the meter.
If no current flows, there is no voltage drop in the resistance R and hence the
voltage at X is the voltage of the battery, or 100 volts. Suppose, however, the
AMMETER-VOLTMETER METHOD 33
meter nas a resistance of 1000 ohms. The current flowing is given by Ohm’s
law
Hs IGOR
= 100 + (10,000 + 1000)
= T1000 = .0091 ampere or 9.1 milliamperes.
This current thrcugh the 19,000-chm resistance R (which may be the internal
resistance of the battery E (Section 49) causes a voltage drop across this
resistance of I X R = .0091 < 10,000 = 91 volts.
The voltage actually recorded on the meter, then, 1s the difference between
the battery voltage and the drop across the resistance R, or
voltage at X = E — (I X R) = 100 — 91 = 9.0 volts.
If, however, the meter is a high-resistance meter, say 1000 ohms per volt,
that is, 100,000 ohms for a meter designed to read 100 volts, the current taken
from the battery would be
I = £/R = 0.00091 ampere
and the 7R drop across the resistance F would be only
H = IR = (0.00091 x 10,000)
9.1 volts
and the voltage read at X would be 100 — 9.1 volts or 91.9 volts.
In other words the high-resistance voltmeter gives a reading much nearer
the open-circuit or no-load voltage desired.
31. Ammeter-voltmeter method of measuring resistance.—
- The example in the above section gives a clue to a good method of
measuring the resistance of a device. The method consists in
measuring the voltage across the device when a measured current
flows through it. If the resistance of the voltmeter is high com-
pared to that of the device, its
own resistance need not be con-
sidered, and the inclusion of the
ammeter into the circuit need not
be taken into account aside from
exceptional cases.
Fic. 29.—Ammeter-voltmeter meth-
Example 6-2. Consider the circuit od of measuring resistance.
in Fig. 29. A voltmeter across the
device whose resistance ig unknown reads 75 volts, and the current meter
(1) indicates a current of 0.05 ampere. What is the unknown resistance?
R = E/I = 75 + 0.05 = 1500 ohms.
34 OHM’S LAW
32. Voltmeter method of measuring resistance.—If the resist-
ance of a voltmeter is known, a resistance can be measured by
its use and a battery. Weston Models 301 and 280 each have
resistance of 62 ohms per volt, so that a 50-volt meter would have
a resistance of 3100 ohms. Take two readings, one of the battery
alone and one of the battery in series with the unknown resistance.
Then the desired resistance may be found from
Ee
I = — — ] Fenn
(7 )
where KE, = Voltage of battery alone;
E Voltage across battery and resistance;
Rm = Resistance of meter.
I
33. Use of low-resistance voltmeter and milliammeter in high-
resistance circuits.—A high-resistance voltmeter is expensive but
is necessary when the voltage output of a socket power device for
supplying plate voltages to a radio set, or any other device which
has a high resistance, is to be measured. The current taken by
the meter is so low that the resistance drop, caused by this current
flowing through the internal resistance of the device, is small com-
pared with the voltage being measured.
A method of using a low-resistance voltmeter and a millia-
ammeter is shown in Fig. 30. Suppose the milliammeter is placed
in series with the output resistance of the device across which the
voltage is to be measured. Suppose the current without the volt-
meter attached is J and the current with the voltmeter attached
is I’. Let E’ be the voltage indicated by the voltmeter when the
key is pressed. The resistance in both cases is the ratio of the
Key voltage and current. Thus,
I = current without voltmeter;
R E = voltage without voltmeter;
I’ = current with voltmeter;
eT E” = voltage with voltmeter;
Fie. 30.—A means of avoiding
/
the use of an expensive high R= E = Er
resistance voltmeter. I Ta
RESISTANCE MEASUREMENT 30
whence FH = desired voltage
i
= BE’ x —
Xx 7
E’
and k= TT
34. Resistance measurement.—Resistances are usually meas-
ured by what is known as the “ comparison” method, that is, by
comparing them with resistance units whose values are known.
For example, we might measure the current through an unknown
resistance, 1, as in Fig. 31, and then adjust a variable resistance,
i
Fig. 31.—Measuring resistance by Fig. 32.—Wheatstone bridge for
comparison. measuring resistance.
Re, whose values may be read directly until the same current flows
under the same e.m.f. The two resistances are then equal in value.
The usual laboratory method employs a Wheatstone bridge.
In diagrammatic form it is represented in Fig. 32, in which R; and
R2 are fixed resistances whose values are known, R3 is the unknown
resistance whose value is desired, and 4 is a variable resistance
to which the unknown is compared and the values of which are
known. The method is as follows. A current is led into the
“bridge ” arrangement of resistances at the points A and B and
36 OHM’S LAW
a sensitive current indicating meter placed at the points X and Y.
The values of R1, Re, and R4 are adjusted until the meter, g,
shows that no current flows through it, that is, there is no differ-
ence in voltage between the two points X and Y which would force
current through the meter. In other words X and Y are at the
same voltage.
The total current J divides at A and flows into the “ arms ” of
the bridge forming the currents J; through Ri and Re and [2
through R3 and R4. If there is no potential difference between
X and Y, the voltage drop along R: is equal to the voltage drop
along R3.
Thus IR, = I2ks (1)
Similarly I, R2 = [2h (2)
R R
Dividing (1) by (2) a = oF (3)
Suppose FR; and Re are equalin value. Then equation (3) becomes
1 = R3/ Rs
or
R3 = Ra,
and to find the value of the unknown resistance Rs we need only
adjust R4 (whose values are known) until no current flows through
the meter. Then the two resistances are equal. Suppose, how-
ever, that the unknown resistance is much larger than any value
we can obtain by adjusting Ry. For example, let it be ten times
as large. Then it is only necessary to make R; ten times as large
as R2 when (3) becomes
Ki) ha = hs) ha 10
R3 10° Ra,
and it is only necessary to adjust Rs until no current flows through
the meter and to multiply the resistance of this standard Rs by
10 to get the value of the unknown resistor R3.
RESISTANCE MEASUREMENT 37
The resistances R; and Rez are called the ratio arms; R4, the
standard resistance, is usually a resistance “ box,” that is, a box
in which are a series of resistance units accurately measured and
equipped with switch arms so that any value of resistance may be
Flexible
Connection
12 O<Phones) 11
Fig. 33.—A single slide wire bridge.
obtained. A simple “slide wire ”’ bridge is shown in Fig. 33. The
unknown and known resistances are compared by means of a
slider on a piece of resistance wire. The relative lengths of the
wire provide the ratio arms R; and Ro.
CHAPTER III
PRODUCTION OF CURRENT
ELECTRICAL energy does not exist in nature in a form useful to
man. It must be transformed from some other form of energy.
For example, the mechanical energy of a motor, or steam engine,
may be transformed into electrical energy by means of a generator.
The commonest sources of current useful to radio workers are
the battery and the generator. The battery is a device which
converts chemical energy into electrical energy; the generator
uses up mechanical energy with the same result.
35. Batteries.—A battery is made up of one or more units
called cells. The essentials of the cell are three: two conductors
called electrodes, usually of different materials, and a chemical
solution known as the electrolyte which acts upon one of the
electrodes more than it does upon the other. In this action, one
of the electrodes is usually “‘ eaten up,” and when this conductor,
usually a metal, is gone, the battery is exhausted; it must be
thrown away or the metal replaced. If the metal can be replaced
by sending a current through the cell from some outside source,
that is, by reversing the process through which the cell was
exhausted, the cell is known as a secondary or storage cell. If the
cell must be thrown away when one of the electrodes is “eaten
up,” it is called a primary cell. The dry cell is a well-known
example.
Experiment 1-3. If a plate of copper and a plate of zinc are placed in
dilute sulphuric acid and a sensitive meter is placed across the terminals as
shown in Fig. 34, a voltage of definite polarity will be indicated. The positive
terminal of the voltmeter must be placed on the copper plate in order that the
meter needle shall move in the proper direction. The copper plate is therefore
positive; the zine is negative. If a heavy external wire is attached to the
plates, a current flows; the zinc is slowly dissolved, hydrogen bubbles appear
; 38
ELECTROLYSIS 39
at the copper plate, and finally the voltage of the cell falls off. Other com-
binations of metals should be tried.
The number of combinations of conductors and solutions that
will make up a primary cell is very large; only a few of them are
useful. Some deliver but small currents and low voltages, others
give off noxious fumes, others do not last long enough to be
practical.
The e.m.f. of such a cell depends upon the nature of the elec-
trolyte and the materials from which the plates or electrodes are
KN
an a
Fig. 34.—A simple primary Fig. 35.—An experiment in
cell. electrolysis.
made. Copper and zinc plates immersed in a solution of dilute
sulphuric acid will give an e.m.f. of about 1 volt regardless of
the size of plates or their distance apart. Zinc and carbon plates
in chromic acid give an e.m.f. of about 2 volts.
Until the plates are connected externally by a conductor there
is a difference of electrical potential existing between the two elec-
trodes but no flow of current. This voltage is known as the e.m_f.
of the cell. When the plates are connected and the cell is put to
work the destruction of the zinc begins. When the zinc is all
destroyed the cell is dead.
36. Electrolysis.—The appearance of hydrogen bubbles at the
copper electrode forms the basis of an interesting experiment which
is illustrated in Fig. 35.
Experiment 2-3. Dip platinum electrodes into a solution of sulphuric acid
and pass a current through them from a battery of about 10 or 12 volts.
40 PRODUCTION OF CURRENT
Hydrogen will be evolved at the electrode attached to the negative battery
terminal and oxygen at the other. These gases will exert sufficient pressure
to force the solution out of the inverted test-tubes. If the volumes of the
two gases are measured it will be found that the hydrogen always occupies
just twice the volume required by the oxygen—which is one of the best proofs
we have that water is made of two atoms of hydrogen combined with one of
oxygen, whence arises the familiar chemical formula for water, H.O.
This phenomenon in which a substance is broken down by
dissociation and then under the action of an electric current is
deposited on one of the electrodes is known as electrolysis. In
this case if the solution had been copper sulphate, copper would
have been deposited on the negative electrode. In the practice of
electroplating, metal from one electrode is deposited on another,
the strength of the solu-
tion, which is a solution
of the metal to be de-
Brass Binding
ea posited, remaining un-
| IL. changed.
fauna Fabel 37. Common dry
4_ Paste of Sal- ammoniac
poraus Paver cell.—The common dry
cell is illustrated in Fig.
1_ Manganese Dioxide-Powdered Carbon 36. ‘The zinc container
is the negative elec-
H- Carbon Rod trode, the carbon rod
in the center is the
positive electrode. The
electrolyte is a mix-
ture of powdered carbon and manganese dioxide moistened
with a solution of salammoniac. The voltage of the cell as
measured by a voltmeter is about 1.5 volts. The ordinary B bat-
teries used in radio are made up of many small 1.5-volt cells con-
nected in series
38. The storage cell.—When the zine container of the dry cell
is eaten up, the cell must be thrown away. In the storage battery
neither of the electrodes is eaten away, but the nature of one plate
is changed and when a current is sent through the battery from
some external source this material is changed back to its original
INTERNAL RESISTANCE 4]
form, so that the cell is said to be “ charged ” and can be used
again.
The storage cell has two electrodes, one of lead and one of lead
peroxide immersed in a dilute solution of sulphuric acid. The
usual storage battery is made up of three cells in series, producing
-a voltage of about 6 at the terminals. The positive terminal is
usually marked, either with a red terminal, or a large cross, or in
some other way. It is important to know the polarity of the bat-
tery when charging it. The positive post of the battery should be
connected to the positive post of the charging line.
Experiment 3-3. If two lead plates about 6 inches square are immersed
in a dilute solution of sulphuric acid—say one part of acid to ten of water—and
connected in series with a battery of 6 or 8 volts and an ammeter, a current
will be seen to flow, and the color of the plate attached to the positive terminal
of the battery begins to change and the evolution of hydrogen bubbles at the
other plate will be observed. The current soon decreases. Now replace the
external battery with an incandescent lamp or an electric bell and note that
current flows out of the lead cell and through the light or the bell.
39. Internal resistance.—One might think that an unlimited
current could be secured from a battery if it were short-circuited.
Such is not the case. A very low-resistance ammeter placed across
a dry cell gives a definite reading—it is not unlimited. Something
must be in the circuit which has a resistance greater than that of
the ammeter or the connections. For example a new dry cell will
deliver about 30 amperes through wires of very low resistance.
This something which restricts the current to a limited value is
the internal resistance of the cell. This resistance depends upon
the construction of the cell, its electrode and electrolyte material,
the distance apart of the electrodes, the condition of the cell—
whether new or old. The older the cell the smaller the area of
electrode in contact with the electrolyte and the greater the resist-
ance. The current delivered by a cell is
E
r+R
internal resistance of cell;
external resistance of circuit.
l|
when r
as)
I
42, PRODUCTION OF CURRENT
When a dry cell shows but a few amperes on short-circuit, the
chances are that its zine case is badly eaten up. A voltmeter which
requires very little current for a deflection will still read normal
voltage, 1.5, because the small current through the cell does not
cause appreciable voltage drop. The ammeter, however, draws all
the current the cell can deliver. This current must flow through
the cell as well as through the external circuit and hence there is
a large voltage drop within the cell, leaving little voltage to force
current through the meter.
Cells which have a large internal resistance deliver but smail
currents; low-resistance cells deliver large currents.
When one tests a dry cell with an ammeter, he is actually
ascertaining the condition of the cell by measuring the internal
resistance. When the cell gets old or has been exhausted due to
too heavy currents, its internal resistance becomes high and an
ammeter reads but small current when placed across it.
The storage cell is a very low-resistance device. Its terminal
voltage when fully charged is about 2.1 volts and it has a resistance
of about 0.005 ohm. Placing an ammeter across such a cell is
dangerous. The meter will probably be ruined.
Example 1-3. A dry cell on short-circuit delivers 30 amperes. Its ter-
minal voltage is 1.5 volts. What is its internal resistance?
By Ohm’s law, ih Ss Joep
0) 3 Uy ie
1.5
7 = ——- =| 0:05 ohm,
30
R=100 Example 2-3. The e.m.f. of a battery is 5
volts. When 100 ohms are placed across it the
voltage V falls to 4 volts. What is the internal
resistance of the battery?
In Fig. 37, R = 100 ohms, r = the internal
resistance of the battery, the voltage drop across
the 100 ohms is 4 volts, which leaves one volt
drop in the internal resistance of the cell. The
Fie. 37.—A problem in current through the external 100-ohm resistance
internal resistance. is, by Ohm’s law,
I = 4/100 = .04 ampere.
POLARIZATION 43
This current must also flow through the internal resistance of the battery and
there it causes a voltage drop of one volt.
Ji; = 1b SK ip
1=0.04 xr
r = 1/.04 = 25 ohms.
owt dl
Note that the internal resistance of the cell is represented as
being in series with the voltage and the external resistance. This
is because the current must actually flow through the internal
resistance of all such voltage generators, and hence the resistance
of the device is represented in series with the remainder of the
circuit. Care must be taken in such representations not to place
the voltmeter in the wrong place. In Fig. 37 the voltmeter is
actually placed across the battery and its internal resistance, which
is connected directly to the 100-ohm resistance.
The voltage of the cell on open circuit is its e.m.f. Under load
the voltage falls, and is now labeled as the p.d. (potential differ-
ence). The e.m-f. of high-resistance cells can only be measured by
high-resistance meters, those that take but little current from the
cell. When the voltage of a cell or battery is mentioned its e.m.f.
is assumed unless otherwise labeled.
40. Polarization.—In common with all other cells the zinc-
copper sulphuric acid cell (Fig. 34) suffers from polarization. The
hydrogen bubbles which surround the copper plate decrease its
surface in contact with the active liquid. This increases the cell’s
resistance. In addition, a minute voltage is set up between the
hydrogen and the copper. This voltage is opposite to the useful
voltage and has the same effect upon the usefulness of the battery
as an addition to the cell’s internal resistance.
Various means are taken to overcome the bad effects of polari-
zation. Chemicals may be placed in the cell to supply oxygen,
which will combine with the hydrogen to form water; shaking
the cell may remove the hydrogen bubbles and decrease the resist-
ance of the cell. The manganese dioxide used in dry cell construc-
tion is a depolarizer. But these are only temporary remedies.
Sooner or later the internal resistance of the cell becomes so high
that the cell is no longer useful. A dry cell which has had large
currents taken from it becomes polarized. If it is allowed to
44 PRODUCTION OF CURRENT
stand for a time, the chemicals put into the cell to do away with
the polarization products have the chance to “ catch up ”’ and the
cell is said to have ‘‘ recuperated.” For this reason dry cells and
others which tend to polarize are used only on intermittent service.
Where a constant current is required a different type of cell is
used.
41. Cells in series.—Cells and batteries may be connected to-
gether in several ways. When the positive terminal of one cell is
connected to the negative terminal of the next cell, as in Fig. 38,
they are said to be connected in series. Under these conditions,
the voltage appearing at the two ends of the series of cells is the
-roa te -é-
nema r | ry
| | | | !
e
m2
gor
I
Fria. 38.—Cells in series. Fria. 39.—Cells in parallel.
sum of the individual cell voltages. For example, if we connect
four dry cells in series, each cell having a voltage of 1.5, a volt-
meter across the two ends will register 6 volts. At the same time
the total internal resistance is the sum of the individual resist-
ances and whatever current flows must flow through this resist-
ance.
If the ends of the battery are connected together with a wire of
resistance & the current that will flow may be obtained by Ohm’s
law as
ene ean
Nr he
42. Cells in parallel— When the positive terminal of one cell
connects to the positive terminal of the next cell, and the negative
terminals are connected together, as in Fig. 39, the cells are said
OERSTED’S EXPERIMENT 45
to be connected in parallel. Under these conditions the terminal
voltage of the combination is the same as the terminal voltage of
each cell, but the internal resistance has been divided by the num-
ber of cells, N, and has become r/N.
If the ends of the battery are connected together with a wire
whose resistance is R, the current that will flow is
TEs é
,
N te
Cells may also be connected in what is called a series-parallel
arrangement. In Fig. 40 are
P sets of S cells in series, and
the sets themselves are con-
nected in parallel.
If the battery of cells shown
in Fig. 40 is connected to a
wire whose resistance is R, the
current that will flow is
ete Nei
WW RPRO Sr
where N= PX S.
43. Magnetism.—The second common source of electric cur-
rents is the generator. The magnet is the heart of the generator.
Magnets obey the same laws as the fundamental electrical
charges mentioned in Section 3 in Chapter I. Like magnetic poles
repel each other, unlike poles attract. Two magnets will repel each
other if their North poles are turned toward each other, but will
attract each other with considerable force when a North and a
South pole are brought near.
44. Oersted’s experiment.—A Danish experimenter, Oersted,
in 1819 made the first of a series of discoveries concerning the rela-
tion between electricity and magnetism which resulted in many
modern applications of electricity. His experiment can be repeated
by anyone who has a compass, a battery, and a wire.
f
Fig. 40.—Cells in series-parallel.
46 PRODUCTION OF CURRENT
Oersted’s experiment demonstrated that a conductor carrying
a current of electricity affects a compass needle just as a bar
magnet does. Another experi-
ment will teach us more about
this important phenomenon
known as electromagnetism.
Experiment 4-3. Wind up about
1000 turns of rather fine wire, say
No. 28 d.c.c., on a form about a
half-inch in diameter or with a hole
(Coil into which Iron Core large enough that a bar magnet can
can be placed) be put init easily, similar to that
illustrated in Fig. 41. Connect it
‘S in series with a battery of about 6
volts, and bring a compass near its
Fic. 41.—A aalenci ae two ends. Reverse the current
through the coil, and note change
in direction of the compass needle motion. Note that the two ends of the coil
act toward the compass needle just as a bar magnet would. Determine which
of the coil ends is North and which is South by comparing their action on the
compass with the action produced by the bar magnet whose poles are marked
or by the position the . =
c ithr Mili WWE WW NY
Siege vi My Y. 7 Ne
Experiment 5-3. Scat: FX. Q\ Ze a Ni} i) 'h
ter a quantity of fine : Zoe Ul i
iron filings on a piece of
cardboard about a foot
square. Place one end
= aie
of a bar magnet under ae cS S J
the cardboard, and tap Hl ne me : Ly) yA
the board until the fil- SSSSZZ/U
ings assume a fixed po- "1c. 42.—How the lines of force about a bar
sition, Repeat using magnet are located.
the other end of the bar
magnet, and then with a horseshoe magnet, and finally with the coil of wire
through which a current is flowing. Place the bar magnet parallel to the
cardboard and again scatter iron filings. Repeat with the coil magnet.
Place the bar magnet inside the coil and remove from the cardboard to
such a distance that, with no current flowing in the coil, the iron filings are
affected but little. Connect the battery to the coil and note the increase
in action among the iron filings. Repeat with battery connected and
OERSTED’S EXPERIMENT 47
cardboard removed to such a distance that the filings are not affected.
Then put the bar magnet inside the coil and note increased effect.
Such experiments demonstrate that coils carrying electric cur-
_ rents have much the same properties as iron bars which have been
magnetized. The fact that such a coil or a magnet affects iron
filings or compass needles even though some distance separates
them, shows that in the space between them exists some force.
The iron filings show the general distribution of this force. They
tend to arrange themselves along lines
which concentrate in strength at the two
ends. These lines are called magnetic
lines of force. The concentration points
are called the poles. The space through
which the lines pass is called the magnetic
field.
The number of lines per unit of area,
as in Fig. 43, is called the field intensity
or flux density; and when one line goes
through 1 sq. em. the field strength is one
Gauss. The total number of lines through
any given area is called the flux, and to
find the flux it is only necessary to multi-
- ply the field strength, H, by the area. Thus
Fia. 43.—The number of
Aux 01 A i. lines of force going through
one square inch or other
NMacnetic lines of force may beset up 1M {init of area is knownlas
iron much more easily than in air. The the flux density.
ratio of the numberof lines, under the action
of a given magnetizing force, that exist in iron to those that would
exist in air is called the permeability of the iron. If one line flows
through 1 sq. em. of air and one thousand through the same
area of iron the permeability of the iron is 1000.
The field strength varies inversely as the square of the distance
away from the pole as shown in Fig. 43. The nearer the square
gets to the pole piece the greater the number of lines. As a matter
of fact, halving the distance multiplies the number of lines by four.
The same magnetizing effect can be produced by a strong cur-
48 PRODUCTION OF CURRENT
rent acting through a few turns of wire as by a weak current
through many turns of wire. If the product of turns and amperes,
called the ampere turns, is the same in two cases the magnetizing
effect will be the same.
It is beeause of the increased permeability, and of the con-
sequent increase in flux density, that adding the iron core to the
solenoid, or coil of wire in Experiment 5-3, is so effective. Soft,
iron is most easily magnetized, but loses its magnetism with the
same rapidity. Permanent magnets are made from steel.
45. Faraday’s discovery.—The second important discovery on
the way toward present-day electrical machinery was that of the
celebrated English experimenter, Faraday. The experiment may
be repeated by anyone who has a coil of wire, a bar magnet, and a
sensitive current indicator such as the galvanometer used in
Experiment 6-3.
Experiment 6-8. Construct a galvanometer like that in Fig. 26a and con-
nect to the coil used in Experiment 4-3. Thrust a bar magnet into the coil
quickly, and then remove it. Note the motions of the compass needle. Wind
up another coil with about the same number of turns but of such a diameter
that it can be placed around the first coil easily. Connect the second coil to
a battery in series with a 30-ohm rheostat and place over the first coil and then
remove with a quick motion. Note the compass needle variations.
In all of the above procedure, note the relation between the compass needle
movements and the rapidity with which the various changes are carried out.
Such was Faraday’s experiment. When the bar magnet, or the
coil carrying a current, was motionless, there was no motion of the
needle. When a bar was thrust into the coil the needle moved in
one direction, and when the motion of the bar was reversed the
needle reversed its motion too. There was no metallic connection
between the two coils, or between the coil and the bar magnet—
and yet changing the position of one with respect to the other, or
changing the direction of magnitude or current through one coil
produced some electrical effect in the other circuit.
The facts underlying Faraday’s experiment are these: An elec-—
tric voltage was generated or induced in the coil when the bar
magnet was thrust into it. A voltage of opposite polarity was
generated when the magnet was removed. This voltage sent a
THE ELECTRIC GENERATOR 49
current through the coil and the galvanometer so that the needle
moved. The same explanation holds when two coils are used, one
of them carrying a current, taking the place of the iron bar. When-
ever lines of force are cut by a conductor, a voltage is generated in
that conductor. So long as the conductor moves so that it cuts
the lines, that is, does not move parallel with them, a voltage is set
up. The more lines per second and the more nearly it cuts the lines
at right angles, the greater the voltage. In Fig. 44, so long as the
conductor AB moves in the direction of the arrow, no voltage is
generated because its motion Ane
is parallel to the lines of force. JH Y
But if AB moves up or down Vp : 7)
or in a direction through the Wi =
: B ee |
paper, a voltage will be E=0
E
measured across its terminals.
This phenomenon is known Fic. 44.—So long as the conductor AB
as electromagnetic induction, Moves across the page, no voltage is
and voltages and current in Pa igre amtat es
the conductor are called in- sce orate
duced, and the electrical cir-
cuit in which they flow is usually called the secondary, the induc-
ing circuit is known as the primary.
There is no discovery in electrical science which has been so
important. Almost every application of electricity to modern life
depends upon this discovery of Michael Faraday.
46. The electric generator.—The essentials of a generator of
electricity are first, a conductor, secondly an electric field, and
third a motion of one relative to the other.
A generator converts mechanical energy into electrical energy.
Figure 45 isa simple generator. It consists of a turn of wire which
is mechanically rotated between two magnets. Remembering that
no current due to the induced voltage flows when the conductor
moves parallel with the lines of force, and that the maximum volt-
age is induced when the conductor moves perpendicular to the
lines of force, because at this point the maximum rate of cutting
takes place, let us see what happens as we rotate the coil. ;
In position (a), Fig. 45, the conductor is moving parallel with
50 PRODUCTION OF CURRENT
the field. No voltage is being generated. As the coil moves, how-
ever, it begins to cut the lines at a greater and greater angle until
finally at (b) it is moving perpendicular to the lines and the voltage
is a maximum. Now as the coil continues to move, the part A-B
instead of moving upward across the lines of force moves downward
across them. The induced voltage then has reversed its polarity
and is increasing toward another maximum position, after which it
returns to its original position, where the induced voltage is again
zero.
In one complete revolution of the conductor there are two posi-
tions at which there is no induced voltage and hence no current in
Fig. 45.—In (a) the conductor AB is moving parallel or along the lines of
force. In (6) the conductor is moving across or at right angles to the lines.
At (a) the generated voltage is zero; at (b) it is a maximum.
the external circuit, and two in which the voltage is at maximum,
although in opposite directions. At intermediate positions, the
_ voltage has an intermediate value.
A complete circle, like a compass, may be divided into 360
degrees. Since the rotating coil moves in a circle, we can label the
positions of the conductor in degrees of rotation instead of positions
(a), (b), ete. At the beginning, (a), it is at 0 degrees—it has not
started to move. Then at (b) it is at right angles to its original
position (a), or it has gone through one-fourth of a complete
revolution, It has therefore passed through one-fourth of 360
WORK DONE BY ATERNATING CURRENT 51
degrees or 90 degrees. When it is again parallel with the lines
of force, it has passed through 180, and when it has gone through
three-fourths of 360, or 270 degrees, the voltage is a maximum but
in an opposite direction, and finally when it reaches its original
position it has passed through a complete circle or 360 degrees.
Let us plot the current induced in the circuit against the degrees
through which the coil has passed. Such a plot is shown in Fig. 46.
When the current is flowing in one direction we call it positive,
when it reverses we call it negative. If the conductor moves at a
uniform rate, say one revolution in 360 seconds, we can plot the
current induced against
time in seconds. 30
One complete revolu- +7
tion is called a cycle. The 3 -
number of cycles per — 2 |
second is known as the a
a 30 60 90 120 150 180 270 360
frequency of the induced
Iternation—>
voltage. The time re- eae =
quired for one complete
revolution is called the ie “=
period. One-half cycle, seer ire
ae Party thes cycle Fie. 46.—Each position of the conductor as
during which the voltage indicated by the arrows corresponds to some
is in the same direction, is yoltage as shown on the graph—called a
called an alternation. “sine wave” of voltage.
Such is the current
produced by an alternating-current, or a.-c., generator. It flows
first in one direction, then in another. The generator, of course,
is a much more complex machine than we have illustrated here.
The magnet is replaced by a heavy iron core covered with wire
in which a direct current flows. This produces a strong uni-
directional magnetic field. The moving coil is also wound on an
iron form and consists of many turns of wire wound in slots.
47. Work done by alternating current.—Some may wonder
whether or not a current that is continually reversing its direc-
tion—never getting anywhere, so to speak—is useful. It is.
Consider a paddle wheel in a stream of water. To the paddle
52 PRODUCTION OF CURRENT
wheel are attached mill stones. Grain is to be ground between the
stones. It matters little whether the water flows continuously
‘turning the millstones in a certain direction, or whether the water
flows first one way and then another. So long as the millstones
turn against each other, grain will be ground. Work will be
done.
The alternating current used to light our homes is usually of
60 cycles, or 120 alternations per second. In some communities
25-cycle current is supplied. In radio installations for use on ship-
board, generators usually produce 500-cycle currents. At the high-
power radio stations of the Radio Corporation of America at Radio
Central, Long Island, are huge alternators which turn out radio
frequencies of 20,000 cycles a second. Smaller generators which
produce frequencies as high as
100,000 per second have been built
but are not in common use.
48. D.-c. generator.—Current is
taken out of a generator by collector
rings. They are illustrated in Fig.
47. If current flowing in a continu-
ous direction is desired, the collector
since Vipul ie ao oR Oa rings are not used, but instead is
an alternating current generator used a device called a commutator.
by collector rings; a commutator 1+ is a switch, or valve, which keeps
serves the same purpose on a the output current flowing in the
direct current machine. same direction by reversing at the
proper time the position of the
wire with respect to the rotating wire. In this manner the current
flows through the external circuit in a single direction, although
the current in the conductor which cuts the lines of force must
reverse each time the conductor passes through 90 degrees and
reverses its direction with respect to the field.
The commutator serves the same purpose as a valve in a pump
which keeps water flowing upward whether the pump handle is
worked down or up. A machine which sends out current which
flows in a given direction is called a direct-current generator, and
naturally, the current is known as direct current, or d.c.
Commutator
(6)
ELECTRICAL POWER 53:
49. Internal resistance.—The generator too has an internal
resistance so that the voltage measured at its terminals differs
when different currents are taken from it. The voltage measured
across the terminals on a meter requiring little current is called
the open-circuit voltage. As the current taken from the generator
increases, this voltage drops.
50. Electrical power.—Throughout the previous discussion we
have spoken of electrical energy in a rather loose way. What do
we mean by energy? What is power? What is their relation to
work?
Energy is the ability to do work. A body may have one of two
kinds of mechanical energy, either potential energy or kinetic
energy. The former is due to the position of the body, the latter
is due to its motion. A heavy ball on top a flag pole has potential
energy because if it falls it can do work, useful or not. It may heat
the ground where it falls, or it may be used to drive a post into the
ground. A cannon ball speeding through the air has energy because
it can do work, useful or otherwise, if it is stopped suddenly. The
target may be heated thereby, converting the kinetic energy pos-
sessed by the ball into heat energy. The amount of damage done
gives the eye a certain measure by which to judge the energy
originally possessed by the cannon ball. This energy was origi-
nally possessed by the powder and was imparted to the ball when
it exploded.
The power required to force a certain current of electricity
through a wire at a voltage of EH volts is the product of the voltage
and the current. Thus
Power in watts equals current in amperes times volts,
or IP = IESE 1B,
A horsepower is 33,000 foot-pounds per minute. It is equal to
746 watts.
All expressions for power involve the factor of time. In other
words, power is the rate of doing work. It requires more power to
accomplish a certain amount of work in a short time than in a
longer time. For example a ton of material raised a foot in the air
represents 2000 foot-pounds of work. If it is accomplished by a
54 PRODUCTION OF CURRENT
crane in 1 second of time it represents an expenditure of 2000 X
60 or 120,000 foot-pounds per minute of power. Since one horse-
power is equal to 33,000 foot-pounds per minute, the crane has a
power of 120,000 + 33,000 or about 3.65 horsepower.
Now if a man raises the ton of material one foot in the air in an
hour’s time by going up a very long and gradual incline, his power
is 2000 + 60 or 33.2 foot-pounds per minute, or roughly one-
thousandth horsepower (0.001 hp.). The amount of work done
in the two cases is the same—the ton of material has been raised
one foot in the air. The rate of doing work has changed.
Example 3-8. A generator is rated at 5 kilowatts (5000 watts) output.
How many amperes can it supply if its voltage is 110 and it has no appreciable
resistance? How many horsepower is this?
Power equals # X I
5000 = 110 x I
I = 5000/1106 = 45.6 amperes.
Horsepower equals 746 watts.
5000/746 = 6.7 horsepower.
51. Power lost in resistance.—According to the law called the
Conservation of Energy, energy can neither be created nor de-
stroyed. It comes from somewhere and goes somewhere. Simi-
larly, all power, which is the rate at which energy is used, must be
accounted for. The energy required to force current through a
resistance must do some work. It cannot disappear. This work
results in heating the resistance. Whenever current flows through
a resistance, heat is generated and the greater the current the
greater the heat. Asa matter of fact the heat is proportional to the
square of the current. If the wire is heated faster than the heat
can be dissipated in heating the surrounding air, the wire melts.
Energy has been supplied to the unit at too great a rate.
A resistor used in a power supply device is rated at so-many
ohms and as capable of dissipating so-many watts. Thus a 1000-
ohm, 20-watt resistor, means that the resistance of the unit is 1000
ohms, and that 20 watts of electrical power can be put into it with-
out danger of burn-out.
EXPRESSIONS FOR POWER 55
Problem 1-3. Electric power is bought at the rate of so-much per kilo-
watt-hour. Suppose your rate is 10 cents per kilowatt-hour. How much
does it cost to run a flat-iron on a 110-volt circuit if it consumes 6 amperes?
Problem 2-3. Assuming that 20 milliamperes direct current flow through
the winding of a loud speaker which has a d.-c. resistance of 1000 ohms, what
amount of heat in watts (J?) must be dissipated? What supplies this power?
Suppose an output device is used which has a resistance of only 200 ohms.
How much power is saved? Is there any other advantage of the latter arrange-
ment you can see?
52. Expressions for power.—Just as there are three ways of
stating Ohm’s law, so there are three ways of stating the relation
between power, volts, amperes, ohms. Thus:
(gee Xe 012). P =) 12 Ro 43) P. = Ree
Example 4-3. A power supply device supplies 180 volts to a power tube
of the 171 type which consumes 20 milliamperes. How much power is taken
from the device? What is the resistance of the tube?
The power supplied is H X J = 180 X 0.02 = 3.6 watts. The resistance
into which this power is fed is equal to P + J? = 3.6 + .0004 = 9000 ohms
or H? + P = 1802 + 3.6 = 32,400 + 3.6 = 9000 ohms.
The maximum current that can pass through one’s body with-
out serious results is 0.01 ampere. The resistance varies with
one’s health and the surface in contact, etc. If the finger tips of
the two hands are dry, the resistance from one hand to the other
is about 50,000 ohms, and thus by Ohm’s law the maximum
voltage that can be safely touched is 500.
Problem 3-3. Assuming that a man can touch with his dry finger tips a
500-volt street car conductor, and that the resistance of his body is 50,000
ohms, how much power is used up in heating the body?
Problem 4-3. A voltage of 110 is to be placed across a circuit whose resist-
ance must be such that 220 watts can be delivered. What is the resistance of
che circuit?
Problem 5-3. How much power is taken from a storage battery of 6 volts
which supplies five quarter-ampere receiving tubes?
Problem 6-3. One milliampere of plate current flows through a 100,000-
ohm resistor. How much power in heat must the resistor be capable of
dissipating?
Problem 7-3. How much current can be sent through a 1000-ohm 20-watt
resistor without danger of burning it up? What voltage is required?
Problem 8-3. Ina plate voltage supply device the voltage-divider has
a total resistance of 5000 ohms. The receiver requires a maximum current of
30 milliamperes. What must be the wattage rating of the resistor?
56 PRODUCTION OF CURRENT
Problem 9-3. A voltage divider has 220 volts across it and a current of
40 milliamperes flows. What is the wattage rating of the resistor? What is
its resistance?
53. Efficiency.—Efficiency is a term that is loosely employed
by nearly everybody. Anything which works is said to be effi-
cient, and one’s efficiency is often confused with his energy—-his
ability to do work whether the work is actually carried out or not.
The term, however, has a very exact meaning when one uses it in
speaking of machines or mechanical or electrical systems of any
kind.
Let us consider a steam engine connected to a dynamo, a com-
bination of machines for transforming mechanical energy into elec-
trical energy. If the steam engine consumes one horsepower (746
watts) and delivers 500 watts of electrical energy, it is said to be
more efficient than if it delivered only 250 watts. Let us consider
two men, one of whom gets a lot of work done in a small amount
of time and with an expenditure of little effort. The other gets the
same amount of work done but with great effort, perhaps flurrying
about from one thing to another instead of tackling his problem in
a straightforward manner. The first man is more efficient. He
wastes less time and energy.
Efficiency, then, is the ratio between useful work or energy or
effect got out of a machine to the total energy or power or effort
put into it. It is expressed in percentage. A machine that is
100 per cent efficient has no losses; there is no friction in its bear-
ings, or, if it is an electrical device, no resistance in its wires.
There are no such machines in use to-day. Efficiency is the ratio
of useful power one gets out of a device to the power put into it.
ful
Enconey useful output E useful output
input output plus losses
Problem 10-3. A generator transmits the greatest amount of power toa
load when the internal resistance of the generator is equal to the resistance of
the load. Assume any resistance and current and calculate the efficiency of
the system. Answer—50 per cent.
Problem 11-3. Seventy-five per cent of the ampere-hours put into a
battery are returned by it on discharge. How many hours must a 100-
ampere-hour battery be charged at a one-ampere rate?
CHAPTER IV
INDUCTANCE
THE experiments of Ohm, Oersted, and Faraday laid the founda-
tion of modern electrical science. The experiments of the last
chapter gave us some idea of what these investigators discovered,
and gave us a background for the following fundamental facts.
54. Coupled circuits.—Consider the two coils P and S in Fig.
48. They are said to be
go through the other.
When P is attached to
a battery and the switch
closed there is a momen-
tary indication of the
needle across S. When
the switch is opened the
needle moves again but
in the opposite direction.
So long as the current
in the primary P is
steady in value and di-
rection, there is no move-
ment of the needie across
the secondary or coils.
If a galvanometer is put
across the secondary of
an ordinary high-ratio
audio transformer, and a
“coupled ” when lines of force from one
Fic. 48.—If the coils P and S are coupled so
that lines of force from P go through S, a
current indicator will show a flow of current
in S when the key in P is closed or opened.
battery is connected across the primary,
the needle of the galvanometer will kick one way when the con-
nection is made, and in
connection is broken.
the opposite direction when the battery
This deflection of the needle indicates a
57
58 INDUCTANCE
momentary flow of current in the secondary coil; this current
flows only when the primary current is changing, i.e., starting or
stopping, not when the primary current is fixed in value or
direction.
55. Lenz’s law.—There are two fundamental facts about this
phenomenon of coupled circuits. The first is that when lines of
force couple two coils together, and some change in these lines
takes place, perhaps due to a change in the current in the circuit
that is producing the lines, a voltage is ‘‘ induced ” in the second
circuit. The second fundamental fact is known as Lenz’s law: this
induced voltage is in such a direction that it opposes the change
in current that produced it.
Thus when the battery is attached to P, lines of force thread
their way across the turns of wire in S. This movement of the lines
of force through S induces a voltage across this coil and a current
flows in the coil and the apparatus connected to it. This current
in the second coil is in such a direction that its field, that is, its
lines of force threading through the primary, induces a counter
voltage in the primary opposite in direction to the battery voltage
across it.
When the battery voltage is broken, the lines of force from the
primary current collapse back on the coil, and, in crossing the
secondary turns in an opposite direction to that taken when
the current in the primary is increasing, induce a voltage in the
secondary in such a direction that it tends to keep the primary
current flowing.
The result is that it takes a longer time to build up the primary
current to its final value at ‘“‘ make” and a longer time for the
current to fall to zero at ‘ break.”
This phenomenon of induced current is of most fundamental
importance. It is the basis of all our modern electrical machinery.
Our motors, our dynamos, our radio signals all are the result of
our ability to produce changes in one circuit by doing something
to another although the two circuits have no metallic connection
whatever.
The fact that current takes longer to reach its final value in a
circuit in which there is a coil of wire indicates that something
SELF-INDUCTANCE 59
about this coil tends to prevent any change in the current. If the
current is zero, this property of the coil tends to prevent any cur-
rent from flowing. If the current already exists, this coil property
tends to prevent either an increase or decrease in this value of
current.
56. Inertia—inductance.—The property of an electrical cir-
cuit which tends to prevent any change in the current flowing is
called its inductance. It has a mechanical analogue in inertia.
A flywheel requires considerable force to get up to speed; and
after it is started it will continue to run for some time after the
driving force is removed. It does not stop suddenly. As a matter
of fact it requires considerable force to stop it, and the more sud-
denly one wants to stop it, the more force he must apply.
Inertia is evident in a mechanical system only when some
change in motion is attempted. It is not the same as friction,
which is always present.
Inductance is a property of an electrical system in which chang-
ing currents are present. It is not to be confused with resistance,
which is always present. Current flowing in a resistance circuit
stops when the driving force (voltage) is removed. If inductance
is added to the circuit, the resistance remaining the same, a longer
time will be required for the current to reach its final value, zero,
when the voltage is removed.
57. Self-inductance.—Inductance is added to a circuit by
winding up a wire into a compact coil. If, for example, 1000 feet
of No. 20 copper wire is strung up on poles, it will have a resistance
of about 10 ohms and the current into it would reach a final value
very soon after a battery were applied. If, however, the wire were
wound up on a spool with an iron core, the time required for the
current to reach its final value would look like the curve in Fig. 49.
Its resistance has not changed; we have merely added in-
ductance.
When the switch is opened a fat spark occurs; this is not true
when the wire is strung up on poles. The current seems to try to
bridge the gap; to keep on flowing. It does not “ want ”’ to stop.
This is because of the inductance of the coil.
A single coil can have inductance and can have a voltage
60 INDUCTANCE
induced across it just as though it were the secondary coil shown
at S in Fig. 48.
When the current starts to go through the coil, lines of force
begin to thread their way out through the coil, thereby cutting
adjacent turns of wire, and according to Lenz’s law inducing in
each turn a voltage in such a direction that it tends to oppose the
building up of the current from the battery. When the battery
connection is broken, these lines of force fall back upon the coil and
=Time for Current to reach
.63 of its final Value
= Time for Current to reach
.63 of its final Value
Current in Amperes
1.0 2.0 3.0
Time in Seconds
Fic, 49.—Current in an inductive circuit does not rise instantaneously to its
maximum value as these curves show.
when cutting the individual turns of wire in the opposite direction
they induce voltages in them which tend to keep the battery cur-
rent flowing.
58. Magnitude of inductance and induced voltage.—The
greater the number of turns of wire in a small space, or the better
the permeability of the core on which the wire is wound, the
greater will be the inductance of the coil and the longer rie
required for the current to reach its final value. The permeability
of air is 1.0; that of iron may be as high as 25,000. This means
MAGNITUDE OF INDUCTANCE AND INDUCED VOLTAGE 61
that the inductance of a given coil may be increased 25,000 times
by winding it on a core of high permeability iron or alloy such as
permalloy in which the permeability may get as high as 100,000.
It is composed of nickel and iron.
The induced voltage across such a coil depends upon the rate
at which the current is changing and the inductance of the coil.
Since the current tends to keep on flowing when an inductive cir-
cuit is broken the voltage across the coil must be in the same diree-
tion as the battery voltage. Thus if a coil has 100 volts from a
battery across it, and the current is suddenly broken, the voltage
at the instant of break across the coil will be 100 plus the addi-
tional induced voltage. Breaking the current into a highly induc-
tive circuit may set up a tremendous
voltage across the coil and a severe
shock can be felt by holding the
ends of the wire at the moment of
break. This is a practical demon-
stration of Lenz’s law.
Example 1-4. In Fig. 50 is an induc-
tance across which is placed a flash lamp
and a battery. The current is regulated so
that insufficient current goes through the
lamp to light it up. Now when the switch
is opened the lamp will suddenly light (and
may burn out) because of the added voltage
across it.
: ; Fig. 50.—An example of Lenz’s
The inductance of a coil depends jy The lamp will light when
upon the number of turns,the manner the key is opened even though
in which the wire is wound, and the insufficient current flows through
material on which it is wound. If the it when the key is closed.
coil is wound on iron, the inductance
will be greater because the lines of force will be concentrated into a
smaller space; more lines per unit area will go through a given
area of coil.
In radio circuits the coils are invariably wound on non-mag-
netic cores. In audio circuits iron is utilized to built up large
inductances in small spaces and with a minimum of copper wire.
62 INDUCTANCE
If an a.-c. voltage is placed across a coil, the current through
the coil is much less than if a d.-c. voltage were placed across it.
This is because of the counter or back voltage induced across the
coil by the effects just described. The result is that a seeming
decrease in voltage across the coil has taken place, although a
voltmeter would indicate that the line voltage was across the coil.
59. The unit of inductance.—When a current change of one
ampere per second produces an induced voltage of one volt, the
inductance is said to be one henry, named from Joseph Henry, an
American experimenter who discovered the phenomenon of electro-
magnetism at the same time as Michael Faraday. Coils added to
circuits for the purpose of increasing the inductance of the circuit
are properly called ‘“ inductors.” In this text the words induc-
tance and inductor are used synonymously—incorrectly, probably.
60. Typical inductances.—The coils used in radio apparatus
vary one inductances of the order of microhenries to very large
ones having over 100
henries in induc-
tance. Broadcast fre-
quency tuning coils
are of the order of
300 microhenries,
=0:394a7N? we and may be from 1
Dera Gg Obes Sire, Conon uh tos inches in diam-
—_9.315a*N 8a + lle .
CEES ee eter wound with from
(a) (b) (c) No. 30 to No. 20 wire
Fia. 51.—Some typical inductances. of from 50 to 100
turns or so. There
are a number of complicated formulas by which one can calculate
the inductance of coils of various forms and sizes. The ones in
Fig. 51 are accurate enough for practical purposes.
These formulas show that the inductance increases as the square
of the number of turns. Thus if a coil of 3 units inductance has its
number of turns doubled, the inductance will have increased four
times or to 12 units. This is true provided there is good ‘“‘coupling ”’
between turns; that is, if the coil is on an iron core this rule is
strictly true, but if the coil is wound on a core of air the rule is
MAGNITUDE OF MUTUAL INDUCTANCE 63
only approximately true. It becomes more nearly a fact the closer
together the turns of wire.
Problem 1-4. A coil like that in Fig. 51a is called a multilayer coil. Such
a coil is wound to have 1000 turns, in a slot 1 inch square. The distance from
the center of the coil to the center of the winding (a in Fig. 51) is 2 inches.
What is the inductance of the coil in microhenries? Remember that the
dimensions given in Fig. 51 are in centimeters and that one inch equals
2.54 cm.
Problem 2-4. A coil like that in Fig. 51b is called a solenoid. It is the
type of coil most often used in radio circuits. The most efficient coil—with
the most inductance for the least resistance—is a coil whose dimensions a and
b are equal. Calculate the inductance of such a coil composed of 60 turns of
wire in a space of 3 inches, the diameter of the coil form being 3 inches.
61. Coupling.—The closer together the two coils, P and S, Fig.
48, the greater the number of the lines of force due to the primary
current that links with the turns of the secondary, and the better
the “‘ coupling ” is said to be. The better the permeability of the
medium in which the lines go, the better the coupling.
The voltage across the secondary of such a two-coil circuit
as that shown in Fig. 48 depends upon the sizes of both coils,
their proximity, the permeability of the medium, and the rate at,
which the primary current changes. All of the factors except
the rate of change of the primary current are grouped together
and called the mutual inductance of the circuit.
The secondary voltage, then, is equal to
M X rate of change of primary current,
where M is the mutual inductance and is rated in henries.
62. Magnitude of mutual inductance.—Formulas in Fig. 51
show that the inductance of a coil depends upon the square of the
number of turns. Doubling the turns increases the inductance
four times. Consider two coils built alike and having the same
inductance. If they are connected together the total inductance
will be equal to that of a single coil of double the turns. In other
words the total inductance of two coils connected “ series aiding ”
will be four times the inductance of a single coil. If the connections
to one coil are reversed, the total inductance will be zero because
64 INDUCTANCE
the lines of force from one coil will encounter the lines of force
from the other coil which are in the opposite direction. The coils
are now connected “ series opposing.”
Consider the series-aiding case, Fig. 52a. The total inductance
is made up of the inductance of coil 1, that of coil 2, the mutual
inductance due to the lines of force from coil 1 which go through
coil 2 and the mutual inductance associated with the lines from
coil 2 which go through coil 1; these two inductances are equal
because the coils are identical.
Thus, Te = In + Ll2 +2M
=21,+2M (because Li = Le)
4 L, (by experiment or measurement)
Ty.
whence M
La=L,+l,+2M
Series Aiding
Series Opposing
(a) (b)
Fic. 52.—Coils may be connected so that their fields aid or ‘‘buck” and
hence so that the total inductance is increased or decreased.
Now if the coupling in both these cases is less than perfect, if
some of the lines from one coil do not link the other—and such is
always the case—the total inductance, Lo, in the series-aiding case
is less than four times the inductance of one coil and in the series-
opposing case is greater than zero. But in any case the total
inductance of two coils of any inductance connected in series-
aiding will be given by L; + Le + 2M = Lg and if they are con-
nected series opposing the resultant inductance will be L; + Le —
2M =I». The following expression involving a new term, the
coefficient of coupling, enables us to predict just what the tote!
MEASUREMENT OF INDUCTANCE 65
inductance in the circuit will be once we know how well the two
coils are coupled.
The coefficient of coupling r = M/V L,Lo.
The coefficient of coupling depends upon the total inductance
in the primary and secondary circuits as well as upon the mutual
inductance between inductances A and B, Fig. 53.
M=0.1 M=0.1
0.1 0.1
T= 10.25 OS SID)
~V.08 X 2.0 ~/(.06 + .08) X 2.0
(a) (d)
Fic. 53.—Examples showing dependence of coefficient of coupling on series
inductance.
The mutual inductance depends upon only the two coils, A
and B, and the coupling between them, or, M = + V I,L2; the
coefficient of coupling between two circuits depends upon the total
inductance in the circuits. The maximum possible value of 7 is
1.0. This is called unity coupling, and Slide Wire
approaches this value in iron-core trans- ra
formers. In air-core coils and transformers
the coupling may be very ‘“ weak,” that
is, of the order of 0.1, and seldom gets
as high as 0.7. In an iron core trans-
former 98 per cent coupling (7 = .98) is
usual.
63. Measurement of inductance.—In-
ductance is usually measured by means of
a Wheatstone bridge (Section 34) just as
resistance ia measured. This is essentially "%% 5+ —l 4 ae sn
; : so adjusted that there is
a method of comparing the unknown in- |, ound in the Shoes
ductance to a known inductance. Re- yi
sistances are used as the ratio arms, as Ly = lye:
in Fig. 54. When there is no sound in
the telephones the inductances are equal if the ratio arms are
66 INDUCTANCE
equal, or if the ratio arms are not equal the unknown inductance
is given by the equation, L, = Ls 5
Mutual inductance is measured on a bridge, or by the following
method: The inductance of the individual coils may be measured
first. Then they are connected series aiding and the total induc-
tance measured. This gives us the formula Li + Lz + 2 M, from
which M can be calculated at once. Of course the same result
will be obtained by connecting the coils series opposing. It is not
even necessary to measure the individual inductances first, pro-
vided we can measure the inductance both series aiding (La) and
then series opposing (Zo). Then
AWE = Iba Ip
bin — bp
4
M =
Problem 3-4. In Fig. 55, when the coils are connected series aiding the
inductance is measured to be 400 microhenries. What
is the mutual inductance?
Problem 4-4. In Fig. 53 (6) calculate the coeffi-
cient of coupling if La is short-circuited, or removed
7=0.6 from the circuit.
Problem 5-4. In a screen grid tube radio-fre-
quency amplifier circuit the primary inductance is 350
microhenries, the secondary is 230 microhenries, the
mutual inductance is 160 microhenries. Calculate the
Fie. 55.—A prob- Coefficient of coupling.
lem in coupled cir- Problem 6-4. In Fig. 55, LZ, = 100 microhenries;
cuits. Ly, = 200 microhenries, and + = 0.6. What is the
mutual inductance? What is the total inductance
(Lo = 11 + Lz +2M)? What is it if L. is reversed (M is negative)?
L, = 100 yh,
L,= 200 zh.
64. The transformer.—A transformer is a device for raising or
lowering the voltage of an a.-c. circuit. It “ transforms ”’ one
voltage into another. It consists ef two windings on an iron core,
as in Fig. 56. The purpose of the iron core is to insure that the
magnetic field set up about the primary will flow through the
secondary coil without loss. What few lines of force do not link
THE TRANSFORMER 67
primary and secondary are called leakage lines and the inductance
associated with them is called leakage inductance.
The primary is attached to an a.-c. line, the secondary to the
load, whether this is a house lighting circuit, a motor, or any other
device which requires electricity.
The lines of force from the continually changing primary alter-
nating current flow through the secondary and induce voltages in
it. A secondary current flows which
increases when the primary current
decreases, and which decreases when
the primary current is increasing.
If there are twice as many second-
ary turns as there are primary
turns, the voltage developed across
the secondary terminals will be frye. 56—A simple transformer.
double that across the primary.
The following formula gives the relation between primary and
secondary turns and the respective voltages:
ees (turns ratio).
Cy uintte
By using the proper ratio of turns, voltages either greater or
less than the primary voltages may be secured at the secondary
terminals.
Example 2-4. A transformer is to connect a 110-volt motor to a 22,000-
volt transmission line. What is the ratio of turns between secondary and
primary?
gD ele
Oe iy Ns
22,000 _ lp 200
Tie a
Nore. This does not give the number of turns in either primary or second-
ary windings. The absolute number of turns depends on several factors; the
ratio of turns depends on the voltages to be encountered.
Problem 7-4. In electric welding a very low voltage is used. What would
be the turns ratio of a transformer to supply a welding plant with 5 volts if it
takes power from the standard: 110-volt circuit?
68 INDUCTANCE
Problem 8-4. A transformer was used in the “old days” of radio when
spark apparatus was standard equipment to step up the 110-volt circuit to
approximately 30,000 volts. What was the turns ratio of the transformer?
Problem 9-4. A primary consists of 200 turns of wire and is connected
to a 110-volt circuit. The secondary feeds a rectifier circuit requiring 550
volts. How many turns will be on the secondary winding?
65. Power in transformer circuits.—Since the transformer does
not add any electricity to the circuit but merely changes or trans-
forms from one voltage to another the electricity that already
exists, the total amount of energy in the circuit must remain the
same, If it were possible to construct a perfect transformer there
would be no loss in power when it is transformed from one voltage
to another. Since power is the product of volts times amperes, an
increase in voltage by means of a transformer must result in a
decrease in current, and vice versa. On the secondary side of a
transformer there cannot be more power than in the primary and
if the transformer is one of high efficiency, the power will be only
slightly less than on the primary side. The product of amperes
times volts remains the same.
Thus the primary power is
BE, I, (1)
and the secondary power is
E.I, (2)
and since there is no loss or gain in power
1 Sd (3)
Lee
when += = i
Teor N (turns ratio),
which shows that the secondary voltage increases as N increases;
the secondary current decreases when N increases.
66. Transformer losses.—Transformers are not perfect. There
is some resistance in both primary and secondary coils. The cur-
rent going through these resistances produces heat, which repre-
sents a certain amount of power lost. All of the lines of force
coming out of the primary coil do not go through the secondary
THE AUTO-TRANSFORMER 69
(the transformer does not have “ unity coupling ”’). Some of the
magnetic field of the primary, therefore, is not used in inducing
currents in the secondary. The iron core—which is a metallic
conductor in the magnetic field of the primary, just as the second-
ary wire is—has currents induced in it and since the iron is a high-
resistance conductor it heats up. All of these losses must be
supplied by the primary source of power, the generator. Large
transformers, however, are very efficient, over 90 per cent of the
input power being transferred to the secondary circuit.
67. The auto-transformer.—It is not necessary for proper trans-
formation of voltage that the primary and secondary windings
shall be distinct. In Fig. 57 is a represen-
tation of what is known as an auto-trans-
former, in which the secondary is part
of the primary. The voltage across the
secondary turns, however, bears the same
relation to that across the primary part as
though there were two separate windings.
The ratio of voltages is the ratio of the p. Ry See ere
number of turns possessed by the second- former. The secondary
ary and primary. can be the entire winding
A transformer is often used when both or part as shown.
alternating and direct currents flow
through a circuit and it is desired to keep the direct current
out of the circuit to which the secondary is attached. No d.-c.
current can go across the transformer when the two windings
are distinct, although the a.-c. voltage variations occurring in the
primary are transferred to the secondary by the effects already
described. If no increase or decrease in voltage is desired the turns
ratio is made unity; that is, the same number of turns will be on
the secondary as on the primary.
Such a case is an output transformer which couples a loud
speaker to a power tube in a power amplifier. The power tube has
considerable direct current flowing in its plate circuit in which the
useful alternating currents also flow. The d.c. is undesirable in the
loud-speaker windings, so a transformer is used to isolate the
speaker from the d.c. of the tube. A good transformer of this typo
70 INDUCTANCE
will transmit all frequencies in the audible range with an efficiency
of about 80 per cent.
Problem 10-4. The line voltage in a certain locality is only 95 5 volts but
a radio set is designed to operate from a 115-volt circuit. A transformer is to
be used like that in Fig. 57. What will be
the ratio of turns?
al ary acts merely as a large inductance
across the line. The current will be
Fic. 58.—An early form of rather small. The energy associated
Toes Reece ce with this current is used in two ways,
urposes ¢ , ° . 5 °
pur . one of which is in heating the trans-
tuning coil.” :
former and its core. The other
part maintains the magnetic field of the primary. This consumes
68. Transformer with open-cir-
cuited secondary.—When no current
is taken from the secondary, the prim-
Fic. 59.—General Radio Company variable inductance.
VARIABLE INDUCTORS 71
no energy from the line because at each reversal of the current the
energy of this field is given back to the circuit.
When the secondary load is put on, however, it begins to draw
current from the secondary and more power is taken from the line
leading to the generator. This additional power is that required
by the load and the loss in primary and secondary resistance.
thd eS
4.5 9
—— -—— =Coils in parallel
= Coils in Series
4 8 Ea a
”
a
£ =]
2 |é
& 3/-s6
0 oa
3 3
E 2
2 25/35 Calibration Curves
= & Type 107G Variometer
= = General Radio Co.
= 2+ =-4
2 =
c c
8 £
3 1.5}—3
&
1 2 LT /
(0) 10 20 30 40 50 60 70 80 90 100
Scale Reading Divisions
Fic. 60.—Calibration of a standard variable inductance.
69. Variable inductors.—A variable inductance may be used
to regulate the current in an a.-c. circuit. This variation in induc-
tance can be secured by means of a slider, as in Fig. 58, or by a
fixed number of turns and a movable iron core. Variations in the
position of the iron rod change the permeability of the core on
which the wire is wound and thereby vary the inductance.
72 INDUCTANCE
For radio-frequency work the variable inductance can take the
form of a variometer, shown in Fig. 59, in which the inductance is
continuously variable from a low value when the two coils are
“bucking ” each other, or are connected series opposing, to the
maximum value where they are connected series aiding. The coils
are always connected in the same manner, but by having one coil
rotate within the other the variations in inductance result. The
calibration of such a variometer is shown in Fig. 60.
70. Effect of current, frequency, etc., on inductance.—A good
air core coil has con-
Measured with 1.0 Volt eftective
applied to Choke Terminals stant inductance at
K=Apparent Inductance of poorly designed Choke all frequencies far
$ =Varation in Inductance with direct Current of :
Samson 30-Henry Choke with one Volt from its natural fre-
A.C. impressed
quency and at all cur-
rents through it. Itcan
be measured at 1000
cycles on a bridge with
the assurance that its
inductance will not be
different at radio fre-
i ee Ie oh |
Milliamperes—D.C. When the current
Fic. 61.—Inductance of iron core coils varies through an iron gore
with d.-c. current because of variation in per- coil changes, the in-
meability. ductance changes be-
cause of the change
in permeability of the core. In order to keep the inductance
more or less constant a small air gap is placed in the core.
Curves indicating the inductance of a choke coil as the current
through it is changed are shown in Fig. 61. The inductance of a
coil to be used where both d.-c. and a.-c. currents are to flow
through it should always be rated by considering the number
of d.-c. amperes that are to flow through it. Thus a coil may
be said to have 30 henries inductance at a d.-c. current of 30
milliamperes. This is the current at which it is supposed to be
used. At other currents it may have more or less inductance.
The effect of leaving a small air gap in the iron core is shown in
Inductance in Henries
EFFECT OF CURRENT, FREQUENCY, ON INDUCTANCE 73
Fig. 62. The air gap decreases the inductance at low values of
d.-c. current but brings it up at high currents and thereby flattens
the curve of inductance vs. d.-c. current.
0
OS OSI 20 253035540 845m 5055
Milliamps D. C.
Fig. 62.—Variation of inductance with air gap.
T—no air gap.
A—average air gap.
B—air gap at one end, 0.01 inch.
C—air gap at both ends, 0.005 inch each.
D—air gap at both ends, 0.0075 inch each.
E—air gap at both ends, 0.01 inch each.
The effective inductance of coils changes greatly at some fre-
quencies and in some circuits. At very high frequencies a coil
may act more like a capacity because of the great number of turns
between each of which exists some capacity. In high frequency
transmitters or receivers, the choke coils to be used may arrive in
the circuit by a cut-and-try method, calculations beforehand
proving worthless.
CHAPTER V
CAPACITY
THERE are two essential electrical quantities in every radio cir-
cuit. These are inductance and capacity. They are represented in
the circuit by the coils and the condensers. Upon their relative sizes
depends the wavelength or frequency to which the receiver or
transmitter is tuned. Resistance is always present too, but the
effort of all radio engineers is to reduce the resistance and to over-
come the losses in power due to its presence, just as mechanical
engineers deplore the share of power wasted in mechanical friction.
71. Capacity.—Inductance has been likened to inertia. In an
alternating-current circuit, it tends to prevent changes in the cur-
rent flowing. Inductance is a property of a circuit; so is capacity.
It is not something one can see, or feel, or hear; one cannot see, or
feel, or hear electricity. We are only aware of it by the work it
does. Inductance in concentrated form is possessed by coils.
Whenever a coil has an alternating current flowing through it, the
inductance is one of its important qualities. Capacity in concen-
trated form exists in condensers. Any two objects which are at
different electrical pressure (e.m.f.) have a certain electrical
capacity. This capacity tends to prevent any change in this
electrical pressure or voltage which exists between these objects.
72. Capacity as a reservoir.—In an electrical circuit, a con-
denser serves the same purpose that the familiar standpipe or
* water tower serves in the water supply system of a city. The water
tower maintains a constant water pressure regardless of the number
of small drains from it, and regardless of the fact that the pumps
filling the tower put water into it in spurts, not in a steady stream
as comes from one’s garden hose.
74
THE CHARGE IN A CONDENSER 75
73. Capacity in a power supply device.—Alternating current
taken from the house lighting wires may be put into a “ rectifier ”’
which cuts off half of the waves, as shown in Fig. 63. These spurts
of current are forced through inductances which delay the rise of
current to its final value on the half cycle in which current flows
from the rectifier. On the other half
cycle, in which no current flows from +N
the rectifier the inductance tends to ae to Rectifier
delay the decay of the current. The
condensers which have been charged [Le oe Nene ere
on the current half cycle discharge Output of Rectifier
during the no-current half cycle. Cur-
rent flowing into these condensers + Fae
charges them to the voltage of the
rectifier system. Then when the Output of First Eliter
rectifier no longer passes current, a rf
the condensers begin to discharge and ESE
0 ee
their voltage falls.
A proper combination of induc-
tance and capacity, called a filter,
will keep a perfectly steady current
flowing out regardless of the fact that
only spurts of current go into the is
system. a “‘B eliminator” varies. The
: half waves from the rectifier
An inductance, then, opposes a are smoothed out by series
change in current; it is like inertia inductances and parallel con-
in a mechanical system. Capacity densers until at /4 it is pure
opposes a change in voltage; it is like direct current.
a reservoir. In an inductive circuit
the current does not reach its maximum value until some time after
the voltage has been applied. In a capacitive circuit the voltage
does not rise to its maximum value until some time after the
current has been flowing in it. A good condenser may keep
charged with electricity for many hours after the charging volt-
age has been disconnected.
74. The charge in a condenser.—A condenser is made up of
two or more conducting plates separated by a non-conductor. The
Qutput of Second Filter
“> dic. 14
0
Final Output
Fic. 63.—How the current in
76 CAPACITY
Leyden jar (Fig. 64) is a good illustration. Filter condensers used
in B eliminators are made up of metal foil separated by waxed
paper. The questions that everyone who looks at a condenser
asks are these: Can a condenser pass an electric current? What is
this capacity possessed by a condenser in concentrated form? How
much capacity is in a condenser? What is the spark that jumps
when a condenser is discharged?
When a condenser is connected to the terminals of a battery, a
45-volt B battery for example, there is a momentary rush of elec-
trons on to one metallic condenser plate and
an exit of electrons from the other. This
constitutes a current flowing into the con-
denser, and if a voltmeter could be placed
across it the voltage would be seen to rise in
much the same manner in which the current
builds up in an inductive circuit. As soon as
the voltage of the condenser is the same as
Tes evden that of the battery, current no longer flows
jar—an early form of into it. It isnow charged, and if the battery
condenser. is disconnected the electrons remain on the
one plate and there is a dearth of electrons
on the other. Now if a wire is connected from one terminal to
the other, these electrons jump across the gap in their effort to
equalize the charge on the two plates.
Once this spark has taken place the condenser is discharged.
So long as the condenser is charged it possesses energy, which
is like the energy possessed by a ball on top a flag pole. The kind
of energy possessed by the ball is potential energy; it is due to the
position of the ball. The energy possessed by the condenser when
charged is also potential energy, due to the strain existing in the
non-conductor. Nothing happens until the condenser discharges,
then it may set fire to a piece of paper, may puncture a hole in a
sheet of glass, or may give some person a severe shock. Thus the
condenser, just as the ball on top the pole, has the ability to do
work—which is our definition of energy.
This is called static electricity, and is the same kind
that produces sparks when we stroke the cat’s back, or rub a comb
QUANTITY OF ELECTRICITY IN A CONDENSER 77
on a coat sleeve, or the kind of electricity that jumps from one
cloud to another on a hot summer day.
75. The quantity of electricity in a condenser.—The quantity
of electricity that rushes into a condenser when it is connected to
a battery is a perfectly definite quantity and can be calculated
or measured. This quantity, Q, rated in coulombs, depends upon
two factors only, the capacity of the condenser and the charging
voltage. The capacity of the condenser depends only upon the
physical make-up of the condenser, that is, (1) the size of the con-
ducting plates, (2) the nature of the non-conductor called the
dielectric, and (3) the distance apart of the plates. The quantity
Q is proportional to both these factors, and may be expressed as
Q (coulombs) = C (capacity) X E (voltage).
The unit of capacity is the farad, named from Michael Faraday,
and is the capacity of the condenser whose voltage is raised 1 volt
when 1 coulomb of electricity is added to it; or vice versa, the
capacity of the condenser to which 1 coulomb of electricity can
be added by an externally applied voltage of 1 volt. This is a
very large unit and in practice engineers have to deal with mil-
lionths of farads, or microfarads. A smaller unit yet, micro-micro-
farad, is used in some radio circuits. This is equal to 10~'” farads.
Another unit is the centimeter of capacity. It is equal to 1.1124
micro-microfarads.
We write the above expression as
Q (coulombs)
C (farads) = RE rom
This expression shows that the capacity of a condenser is the
ratio between the quantity of electricity in it and the voltage
across it.
The third way of stating the relation between capacity, quan-
tity, and voltage, defines the voltage:
Q (quantity)
E (voltage) = Cenes
78 CAPACITY
A discharged condenser, of course, has no electricity, Q, in it
and hence no voltage across it. When it is connected to a battery,
a voltage is built up across the two plates, the value of this voltage
being given at any instant by the ratio between the quantity, Q,
and the capacity, C, of the condenser. The greater the quantity
of electricity stored on the conducting plates, the greater the
voltage. When the battery is removed the quantity of electricity
remains and, of course, a voltage, H, exists between the two plates.
76. Time of charge.—Since an ampere is a rate of flow of cur-
rent, that is, 1 coulomb per second, one can calculate the rate at
which current flows into a condenser provided the quantity, Q, and
the time, t, are known. The amperes before the condenser is
attached to the battery are zero, at the completion of the charging
process the amperes are zero. The average rate, then, is what one
secures from the equation,
Example 1-5. A condenser of 15 microfarads is attached to a 220-volt
circuit. What quantity of electricity flows into it? If it requires 1/200 second
to charge it, what is the average current?
O) SC! Se 18
= 15 kK 10-* x 220
= 3300 x 1076
= .0033 coulomb.
= .66 ampere.
This is also the average rate of discharge if the time of complete discharge
is 1/200 second.
Problem 1-5. What is the capacity of a condenser that holds .0024
coulombs when attached to 220 volts?
Problem 2-5. In a radio circuit is a .0005-mfd. (500 mmfd.) condenser
across a 500-volt source. What quantity of electricity will flow into it?
Problem 3-5. What voltage will be necessary to put 0.012 coulomb inte
the condenser of Problem 1?
ENERGY IN A CONDENSER 79
Problem 4-5. The average charging current in Problem 1 is 2 amperes.
How long will it take to charge the condenser?
Problem 5-5. Suppose the average voltage across a condenser when it is
being discharged is one-half the voltage when fully charged. Connect a
10-ohm resistance across the condenser of Problem 1. What average current
flows? What is the average power used to heat the wire?
Problem 6-5. How long would it take to discharge the condenser in
Problem 5? If the resistance is doubled, what power is used up in heat and
how long will it take to discharge the condenser?
77. Energy in a condenser.—The amount of energy that can
be stored in a condenser in the form of static electricity can be com-
puted from the formula:
energy = 3C E?.
This represents the work done in charging the condenser, and
naturally represents the energy released if the condenser is dis-
charged.
Similarly, the energy in the lines of force about a coil through
which an a.-c. current flows is equal to $ L I?.
The unit of energy or work is the joule. It is the amount
of work required to force one coulomb of electricity through a
one ohm resistance. Thus if a 1l-mfd. condenser is charged to
a voltage of 500, the energy is ;
ae oO se 5007 = 12> joule:
Since power is the rate of doing work, we can find the power
required to charge such a condenser in one second of time by
dividing the above expression by one second. Thus
Ge
Zt
Power in watts =
and if we attach the condenser to a secondary of a 500-volt trans-
former which charges the condenser 120 times a second (60-cycle
current) the power will be = 3 C EH? X 120 provided the con-
denser is permitted to discharge each time. This proviso is
true of the following discussion too.
80 CAPACITY
A general expression for such a problem is
Power = 3 CE?N,
if N is the times per second the condenser is charged and dis-
charged.
Example 2-5. A condenser in a transmitting station has a capacity of
001 mfd. and is charged with a 20,000-volt source. What energy goes into the
condenser and what power is required to charge it 120 times a second (60-cycle
source)?
Energy or work, hy = & Cia
= 4 X .001 X 107® X 20,0002
= 0.2 joule.
Power, IP = lh SS IN
= 5) (CUA
0.2 X 120
24 watts.
ll
Problem 7-5. If the generator in the example above were a 500-cycle
generator, what would be the power taken from it?
Problem 8-5. How much power is required to charge a 1-mfd. condenser
to a voltage of 220, 120 times a second?
Problem 9-5. A transmitting antenna has a capacity of .0005 mfd. It is
desired to transmit 1 kw. of power. To what voltage must the antenna be
charged from a 500-cycle source?
Problem 10-5. Suppose an antenna is to be supplied with 500 watts of
power and that between the charging mechanism and the antenna exists an
efficiency of 30 per cent. The condenser, which discharges into the antenna,
has a capacity of 0.012 mfd., and the generator which charges the condenser
is a 500-cycle machine. A transformer is used to step up the voltage 110 from
the generator to the value required by the condenser. What is the secondary
voltage of the transformer and what is the turns ratio between secondary and
primary? Nore: Remember that a 500-cycle generator charges a condenser
1000 times a second. Answer; 15,200 volts, 131.
78. Electrostatic and electromagnetic fields.—The energy exist-
ing in an inductive circuit is said to exist in the electromagnetic
field surrounding the inductance. This field is made up of lines of
force and can be explored with a compass, or by sprinkling iron
filings on paper as in Fig. 42.
The energy in a condenser is said to exist in the electrostatic
field. This is the locality in which the electrical strain exists, that
CONDENSERS IN A.-C. CIRCUITS 81
is, in the non-conductors in the vicinity of the conducting surfaces
which are charged. This field cannot be explored with any mag-
netic substance, but can be discovered by any form of charged
bodies or any container of static electricity.
Some circuits, an antenna system for example, have both
capacity and inductance and, if properly charged, have both a
magnetic and a static field. Since the wire can be charged to a
very high voltage with respect to earth, considerably energy can
be fed into it.
Frictional electricity, such as that produced by rubbing the
cat’s back, is a form of static electricity. It is this kind of elec-
tricity that is produced in the tank of a truck carrying gasoline
along the road. The gasoline sloshing about in the metallic tank
may raise the voltage of the tank to a considerable degree above the
ground from which it is insulated by the rubber tires. Finally a
spark may pass, and neutralize the charge—but in the process the
tank and driver may be blown to bits. To prevent such accidents,
all gasoline trucks trail an iron chain which connects the tank
electrically with the ground and discharges it as fast as such static
charges are produced.
79. Condensers in a.-c. circuits.—A perfect condenser is one
which is an absolute non-conductor to d.-c. currents—that is, it is
an infinite d.-c. resistance—and one which has no a.-c. resistance.
All the power that is put into it is used in setting up an electro-
static field. Unfortunately all condensers have some a.-c. resistance,
and few have infinite d.-c. resistance. Otherwise a condenser once
charged would keep its charge forever. The time it takes a con-
denser to discharge is proportional to the product of its capacity
and its resistance. This is known as its time constant, and is also
the time required to charge the condenser. The resistance to d.c.
is known as its leakage resistance. In a good condenser this may
be as high as several hundred megohms.
Experiment 1-5. Charge a filter condenser of about 2- to 10-mfd. capacity
by connecting to a 110-volt d.-c. circuit. Then discharge by a heavy wire;
then charge again and allow to stand for a half-hour and discharge. Charge
again and permit to stand for an hour and discharge. T he relative sizes of
spark give an idea of how poor a condenser it may be from the standpoint of
82 CAPACITY
leakage. Then charge and place a 10-megohm resistance across it for a
second or so. Then see if it can
be further discharged by means of
a wire.
A good condenser may retain its
charge for many hours after being
removed from the source of charg-
ing current. The leakage of con-
densers not only takes place across
the insulating material through
Fig. 65.—Leakage of current through which its terminals are brought
the condenser C causes a voltage drop out but through the wax filling,
across the input of the following tube. the container, and through the
dielectric itself.
Example 3-5. In Fig. 65 is a typical coupling device used between tubes
in a resistance-capacity coupled amplifier. A high-voltage battery is used. The
purpose of condenser C Ry R,
is to keep these high d.-c.
voltages from getting to
the grid of the following
tube. If this condenser
has any leakage resist-
ance, a d.-c. current flows
through it and impresses a
voltage on the grid of the
succeeding tube which is Fig. 66.—The circuit equivalent of Fig. 65.
highly detrimental.
Suppose the condenser has a resistance of 100 megohms. What voltage
will be impressed on the grid if the battery has a voltage, Hp, of 200?
100 Meg. Figure 66 represents the circuit. It has
the battery in series with two resistances,
the plate resistance of the tube Rp and
the coupling resistor R. Across Eb and R
is shunted the condenser resistance and the
grid leak in series. The problem is to find
what current flows through this shunt circuit
and what voltage drop this causes across
Fic. 67.—The battery in this the grid leak, Suppose 100 volts appear
figure represents the voltage 2CTS® the series circuit composed of Re and
across Re and Rg in Fig. 65. Rg. This may be as represented in Fig. 67.
One hundred volts across 100 megohms (the
coupling resistance R and the grid leak resistance are negligible in com-
parison to the condenser leakage resistance) produces one microampere of
CONDENSERS IN GENERAL 83
current. This one microampere flowing through the grid leak of one
megohm produces a voltage of 1 volt. This voltage is of a polarity that
is opposite to the C bias and therefore decreases its value. In addition a
certain amount of noise may be developed by the d.-c. current flowing through
this condenser and grid leak. This noise is directly impressed on the input to
the amplifier tube and may assume large values when amplified by succeeding
stages. Condensers used in resistance amplifiers must have a high resistance.
80. Power loss in condensers.—A pure inductance takes no
power from an a.-c. line once the magnetic field is established. If,
however, the coil has resistance, power is wasted in heating it.
Similarly, a perfect condenser, that is, one having no resistance,
wastes no power. If the condenser has resistance—and all have—
power is wasted in heat when it is connected to a source of current,
whether a d.-c. or an a.-c. source. This power in watts is the cur-
rent squared times the resistance.
81. Condenser tests.—If a voltmeter, a battery, and a con-
denser are connected in series a momentary deflection will be noted
if the condenser is good. If the condenser is leaky, a constant
deflection will be noted. If the condenser is fairly large, one or two
microfarads, and no deflection is noted, the condenser is probably
open. If the full battery voltage is read by the meter, the con-
denser is shorted.
If a condenser that has been determined to be not shorted is
placed across a battery and then a pair of phones is placed across
the condenser, a click will indicate that the condenser is good.
The click is the discharge of the condenser through the phones.
82. Condensers in General.—A condenser, then, is any object
with two or more conducting plates separated by a non-conductor
called the dielectric. A cloud filled with moisture is a good con-
ductor. So isthe earth. The cloud may be a mile or so above the
earth and subjected to all sorts of electrical differences of potential.
It may become charged with respect to the earth. When this
charge becomes great enough to jump the gap, a spark passes, and —
the condenser is discharged. This is known to us as lightning.
Small discharges between parts of a cloud or between two clouds
may cause only smallsparks and are made known to us by “ static.”
The earth is considered as being at zero potential. All objects
not connected to earth have a higher voltage than the earth, and
84 CAPACITY
hence any object has a capacity with respect to earth. This capac-
ity will conduct electricity just as a one microfarad by-pass con-
denser in a radio receiver will, and may cause considerable embar-
rassment to the radio engineer or the experimenter who does not
take it into consideration in his calculations.
In radio circuits the conducting plates of a condenser may be
aluminum or brass or tin foil, depending upon the service which
the condenser is to fill. The insulating plates, called the dielectric,
may be air, oil, mica, coated paper coated with beeswax or other
insulating compound. The actual capacity of the condenser may
be fixed, or it may be variable.
83. The nature of the dielectric—If two square metal plates
10 em. on a side are suspended in air and about 1 mm. from each
other, the capacity will be about 88.5 mmfd. If a sheet of mica is
between the plates the capacity will be increased about eight times.
Other substances will give different values of the capacity. Each
substance, in fact, will give a certain value of capacity depending
upon what is called the dielectric constant of the substance. The
table below gives the value of K, the dielectric constant, of several
substances.
This factor K has nothing to do with the ability of a substance
to withstand high voltages without puncturing. Such ability dif-
fers not only with the substance but also with the condition of the
substance at the time the voltage is applied, that is, the percentage
moisture present, the pressure to which it is subjected, etc. Mica
for example will withstand much greater voltages than paraffined
paper.
* Dielectric . Dielectric
EMSS Constant K Sa Constant K
INIAP Sorc OR REE tee 1 Para ihe eee 2
IB AUKE ILE CH asc che hucio eae 4to8 Porcelainw ee DELOIG
Wellloid's as cece se 4 to 16 Rubber. oe eee 2 to 34
(CHES). oa QS aene eae 4 to 10 PYTOX hotest rae 5.4
IVITCOMPMENE TS spices ate ws aan 3) WO 7 Shellactaasqeeee nee m3, (5)
OUACASTOM jc ays 5 2s os oS 4.7 Varnished cambric... . 4
Oil, transformer....... 2.2 W 00d 252. acted 2 to8
IOAIEP. oo 10 60 2 to 4
,ea—Kre_— TT SS
CONDENSER CAPACITY FORMULAS 85
84. Sizes of radio condensers.—The farad is a very large unit.
In radio work the capacities used vary from a few million millionths
of a farad to several millionths of farads. A millionth of a farad is
known as a microfarad, and a millionth of this value is called a
micro-microfarad. These are related as shown below.
One farad equals one million mfd. and one million million
mmfd.
One mfd. equals one millionth farad and one million mmfd.
One mmfd. equals one million millionth farad, and one mil-
lionth mfd.
Sometimes the centimeter is used as a unit of capacity. It is
equal to 1.1124 micro-microfarads.
85. Condenser capacity formulas.—Formulas have been worked
out by which it is possible to compute the capacity of condensers
of various forms. For example the capacity of two flat conducting
plates separated by a non-conductor may be computed from the
formula:
C 885 <A XK
NO
where C = capacity in microfarads;
A = area of the metallic plate in square centimeters;
K = dielectric constant of the non-conductor;
d = thickness of dielectric in centimeters.
Or
Le
C = 0.0885
where C = capacity in micro-microfarads;
d = thickness in centimeters;
A = area in square centimeters;
K = dielectric constant.
Formulas for other types of condensers may be found in the
Bureau of Standards Bulletin 74, Radio Instruments and Measure-
ments, page 235.
86 CAPACITY
In receiving sets the tuning condensers are variable. The
important function of separating one station from another is per-
formed by changing the capacity of the condenser, which is called
tuning the circuit. The tuning condensers have air as the dielec-
tric and plates of various metals, usually brass or aluminum. The
capacities range from 25 mmfd. to 100 mmfd. in a short-wave
(high-frequency) receiver to 500 mmfd. for broadcast frequency
receivers. Receivers for the long waves used in transoceanic com-
munications are much larger. The above values of capacity may
be written as .0001 and .0005 mfd. It is probably easier to express
small capacities in micro-microfarads rather than in one of the
larger units. Whenever an easy path for an alternating current is
needed a fixed condenser is used. It is called a by-pass condenser.
Example 4-5. How many plates 16 < 20 cm. in area and separated by
paraffined paper (K = 2.1) .005 cm. thick are required for a condenser of
24 mfd.?
_ 885 AK
10d
1-885 X16 x 20.21
= 10 x .005
= 0.0119 mfd. capacity per plate.
24
Number of plates = — 2200;
eee tee 0119 :
Problem 11-5. Express in micro-microfarads: zoo farad, 1 mfd., 0.00025
mfd. Express in farads: 500 mmfd., 0.01 mfd., 745 mfd. Express in micro-
farads: 1 farad, 500 mmfd., 0.01 mmfd.
Problem 12-5. What is the capacity of two square plates suspended in
air 0.1 mm. apart, if the plates are 10 em. on a side?
Problem 13-5. A variable tuning condenser has a capacity of 0.001 when
air is used as dielectric. Suppose the container is filled with castor oil. What
is the capacity now?
Problem 14-5. Lead foil plates are separated by mica 0.1 mm. thick
having a dielectric constant of 6. What is the capacity of a condenser made
up of 200 pairs of such plates?
Problem 15-5. How many joules of energy can be stored in such a con-
denser as in Problem 14 when 100 volts are impressed across it? How much
power will it take to charge it to this voltage 120 times a second?
CONDENSERS IN SERIES AND PARALLEL 87
86. Condensers in series and parallel— When condensers are
connected in parallel (Fig. 68) the resultant capacity is the sum of
C,
E, c
E, — E,=E E
Fic. 68.—Condensers in parallel. oe
the individual capacities. When they are in series (Fig. 69) the
resultant may be found as below:
Conia a Gi oF Ce a C3 Le reset
1
Cees ae l F 1 is 1
Gr C2 80s >
The resultant of two capacities in series is
C Cy X Ce
fes — :
Series Cr ae Ce
E, E, C
C, Gp ee
c E
E=E£,+£, aC ,Cs
Where C Cats
Fic. 69.—Condensers in series. E=E
In the parallel case the same voltage is across each capacity. In
the series case the voltage varies inversely as the capacity. If two
equal capacities are in series half the total a.c. voltage is impressed
across each condenser. If one condenser has half the capacity of
the other, that one will have two-thirds the total voltage whereas
the other will have one-third the total voltage across it.
The resultant capacity of several condensers in parallel is
always greater than any single capacity; in series the resultant is
88 CAPACITY
less than the capacity of the smallest of the group. In series the
voltage across each condenser varies inversely as its capacity and
is always less than the total voltage across the combination. When
a condenser is placed in an a.-c. circuit whose voltage is more
than the condenser can tolerate, it is only necessary to use two or
more condensers in series so that the voltage across individual
members of the group is below the danger limit, and so that the
total capacity in the circuit is the value desired.
When condensers are used across d.-c. circuits, as in filter
circuits, the d.-c. resistance of the condenser becomes of impor-
tance. If two condensers in series across a certain d.-c. voltage
have equal capacities but different d.-c. resistances, the voltage
drop across the two condensers will differ. The voltage across
one of them may be sufficient to destroy it.
Example 5-5. In a radio circuit a .0005-mfd. variable condenser is avail-
able but the circuit calls for a .00035-mfd. condenser. What fixed condenser
may be used to reduce the maximum capacity in the circuit to the proper
value? How shall it be connected?
Solution. Since the total capacity is to be reduced, the fixed condenser
must be connected in series with the variable condenser. The total capacity
is given as 0.00035 mfd. This is equal, from the above equation, to
C Bes
00035 = 21 XC
C1 +C,
a .0005 X C2
.0005 + C2
whence C, = .001166 or 1166 mmfd.
Problem 16-5. Ina transmitter a ‘‘ blocking ’’ condenser is needed whose
capacity must be approximately 0.001 mfd. The voltage it must stand is 1000.
A 0.002 condenser is available which can stand only 400 volts. What size
condenser must be used in series so that no more than 400 volts is across the
0.002 condenser? (Neglect resistance of condensers.)
Problem 17-5. Whenever the capacity of a tuning circuit is quadrupled,
the inductance remaining constant, the wavelength is doubled and the fre-
quency is halved. A broadcast receiver tunes to 600 meters (500 ke.) with a
500-mmfd. condenser. What maximum capacity must be put in parallel with
this condenser in order to receive 800-meter (375-kc.) radio compass signals
and ship-to-shore traffic on 2,200 meters? Ans. 390 mmfd., 6250 mmfd.
CHAPTER VI
PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
Tue two kinds of current in common use are: direct currents
(d.c.) which have a more or less constant value and which flow in
the same direction all the time; and alternating currents (a.c.)
in which not only is the magnitude constantly changing but the
direction also.
87. Definitions used in a.-c. circuits—When the voltage (or
current) has started from zero, risen to its maximum value in one
direction, decreased to zero and risen to maximum value in the
opposite direction and finally come back to its starting value and
point, zero, it is said to have completed a cycle. Ordinary house-
lighting current which has a frequency of 60 cycles per second goes
through this cyclic change in magnitude and direction 60 times a
second. The frequency is the number of times a second a cycle is
completed. An alternation is half a cycle; that is, when the voltage
has gone from zero to zero through one maximum it is said to have
completed one alternation. In 60-cycle circuits there are 120 alter-
nations per second. In a.-c. circuits we must consider the element
of time; in d.-c. circuits time does not enter; the magnitude
of the current is constant.
Alternating currents exist of nearly all ranges of frequencies.
Sixty cycles is the common power frequency; tones generated by
audio-frequency oscillators for testing purposes may go from almost
zero frequency to as high as the human ear can hear, that is, about
13,000 cycles per second depending upon the person. Electric
waves of frequences as low as 15,000 cycles exist. They are gen-
erated in the long wave high-power radio stations carrying en
transoceanic communication. From that value radio frequencies
89
90 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
exist up to about 30,000,000 cycles per second. This corresponds
to a wavelength range of from 20,000 meters to 10 meters.
88. Instantaneous value of alternating current.—Since the
voltage (or current) is continually changing in value it becomes
expedient to provide a means of knowing what this value is at
any time.
Time :
Fic. 70.—As the vector E rotates, the circle moves to the right. A piece of
crayon attached to the end of the vector would trace out a curve similar to
the wavy line. When the vector has the position as shown in the first small
circle, the instantaneous value e is zero. At other times the instantaneous
voltage has some other value.
Let us consider the circle of Fig. 70 which is moved to the right
ata constant rate. Within the circle is a rotating arm, representing
the motion of the rotating part of an a.-c. generator, as well as the
voltage, H, it produces. It rotates at a constant eoeed such that
one full rotation—one cycle—is completed in the time it takes the
circle to move to the right a distance equal to the diameter Now
when the arm is perpendicular to its starting position, rhe dinate
INSTANTANEOUS VALUE OF ALTERNATING CURRENT 91
has completed one-quarter of its movement ; when the arm is
pointing in a direction opposite to its starting position the circle
has moved through one-half of its motion, and so on. Now let us
picture what the end point of the arm would trace out if we
attached a crayon to it and let it go through its motion as the
circle is moved to the right. Such a tracing will be an accurate
representation of an alternating current or voltage.
In Fig. 70 the height of the arm above the horizontal axis, its
starting position, represents the value of the voltage at the instant
the generator winding is at the position in its cycle corresponding
to the position of the rotating arm whose length is E. The posi-
tion of the arm at any point is known as its phase. Since a cycle
is represented by a complete circle of 360 degrees (360°), when the
arm is vertical we speak of its position at the 90° phase. Now the
height of this arm from the horizontal starting position is the value
of the a.-c. voltage at that position or phase or instant of time.
At 90 degrees this height is equal to the length of the arm itself, or
the a.-c. voltage is at its maximum value; at any other point in
that half-cycle or alternation the voltage is less than this value.
Now it is handy to have something to which to compare the
value of the voltage at any phase, known as the instantaneous
value—because this value is only temporary due to rotation of the
alternator armature. This basis of reference is the maximum or
peak value. The instantaneous value is always rated by stating
its magnitude with respect to the maximum value. Fortunately
there is a factor which relates the height of the arm representing
the instantaneous value and its length, or maximum value. This
factor is known as the sine of the angle between the arm and the
vertical line. Knowing the maximum value of an a.-c. voltage—
the length of the rotating arm in Fig. 70—and the phase, or the
angle through which the arm has rotated, to determine the instan-
taneous value of the voltage we need only multiply the maximum
value by the sine of this angle which we may look up in a table
made out for such a purpose. For example, the functions of several
angles are given below. The angle itself (which is a means of
expressing the time that has elapsed since the alternator arm
started rotating) is called the phase angle.
92 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
Angle
Degrees
15
30
45
60
90
110
135
175
180
220
270
300
88a. Triangle functions.—Considering the angle between CB
and AB in Fig. 71 labeled as 6 (the Greek letter theta), the side
A AC is called the opposite side, CB is
called the adjacent side. Then these rela
tions hold:
AC.
— = tan 0 -
CB an (1) or AC = CBtan @
Aue oe :
c yee sin 6 (2) or AC = ABsin 0
AC
cp land
FG o, B
ABO —=cosé (3) or CB = AB cos 0
CB cosa AB
AB
. 71.—Tri Ti > : '
a be ie ees Knowing any two of the three functions
of the right-angled triangle and the angle
involved, the other function may be found by means of the table
in Section 88.
Notre. The terms “sine,” “ cosine,” ete
., are called the
functions of the angles.
MEANS OF EXPRESSING INSTANTANEOUS VALUES 93
The function of angles greater than 90° may be found from.
function (NV X 90° + A) = function A, if N is even, e.g., sin 210°
= sin (2 X 90° + 30°) = sin30°. Function (N x 90° + A) =
cofunction A if N is odd, e.g., sin 120° = sin (1 x 90° +30°)
=) ¢0s:30 .
89. Means of expressing instantaneous values—We may
express the instantaneous value of an a.-c. voltage or current as
follows:
e=EHsné@ or 1=T/sin8@,
where e or? = the instantaneous value;
E or I = the maximum value;
6
the phase angle in degrees.
=20 Sin 90°
=20x1
=20
=20 Sin 45°
=20X.707
=14.
6=45° 0=90°
@=E Sind
=20 Sinw
=20 Sin 27°
=9.0
€=E Sing
=20 Sin 135°
= 20.707
=14.,
§=135° 6=333
Fic. 72.—At various times in the cycle the instantaneous value of the vector
E is as shown in these vector diagrams.
Small letters denote instantaneous values, capital letters denote
maximum values.
Let us look at Fig. 72 which represents the rotating arm EH and
94 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
the vertical height e in four typical cases, 45°, 90°, 135°, and
333°. The actual height compared to the maximum length of the
arm E may be calculated by means of the above table and formulas.
Since the sine of an angle of 0 degrees is zero, the instantaneous
value of the voltage at 0 phase is zero; since the sine of an angle
of 90° is 1, the instantaneous value of the voltage at this point in
the cycle is equal to the maximum value; and so on.
The three methods of representing an a.-c. voltage or current
are:
1. By a graphical illustration such as Fig. 70, called a sine
wave.
2. By an equation, such as
e= Esin@ or «=/sin@.
3. By the pictures shown in Fig. 72, known as vector diagrams.
Such a line as # in Fig. 73 which moves about a circle is called
avector; the vertical distances of its
Componente end point from the horizontal axis is
called its vertical component. The
angle 6 between the horizontal and the
vector is called the phase angle; the
value of the vertical component may be
found by multiplying the maximum
value EK by the sine of the phase angle.
Phase Angle @
Fie. 73.— The maximum
value of the vector is H; the
instantaneous value is e.
Example 1-6. Represent a voltage whose
maximum value is 20 volts. Using graph or
cross-section paper twelve divisions to the
right equals one alternation and ten divisions
vertically up and down from our horizontal line represents the maximum
values of the voltage. The voltage starts at zero, increases to a maximum
at 90° or six divisions, then decreases to zero at twelve divisions, ete. What
is its value at other times? We can tell by using our table of sines in Sec-
tion 88. At the 30° phase the instantaneous value, or the height of rotating
arm above the time axis, is e = Hsin 30° = 20 X .5 =10 volts. Other
instantaneous values can be found similarly and the entire sine wave plotted
similar to Fig. 70.
Lay off on cross-section paper a length say 20 divisions equal to the maxi-
mum value of the voltage. Then using this value as a radius draw a circle.
EFFECTIVE VALUE 95
Then the instantaneous value at any time in the cycle can be found by measur-
ing the vertical distance of this point on the circle to the horizontal axis.
Thus at the 30° phase the vertical distance is 10 because sin SO = A ‘MWlos
illustrated the vector diagram.
A voltage of maximum value 20 may be represented mathematically as
e = 20 sin @
Problem 1-6. The maximum value of an alternating voltage is 110. What
is its value at the following phases: 30°? 60°? 110°? 180°? 270°? 300°?
360°?
Problem 2-6. The instantaneous value of an alternating voltage is 250
volts at 35°. What is its maximum value? What is its value at 135°?
Problem 3-6. The instantaneous value of an alternating voltage is 400
volts at 75°. Plot to some convenient scale its sine wave.
Note. In all this discussion voltages or currents can be spoken of with the
same laws in mind. Thus the form of a sine wave of current looks exactly
like the sine wave of voltage with the same maximum value. The vector
diagram looks the same because it is only necessary to label the rotating arm J
instead of H and the mathematical formula reads i = J sin @ instead of
e = Hsin 6. The answers to the above problems will be the same numerically
whether we speak of voltage or current.
A word should be said, too, about the terminology frequently
used in speaking of the voltages and currents in an a.-c. circuit.
Engineers use the expression “ a.-c. voltage ”’ or “‘a.-c. current ”’
for simplicity, not stopping to think that such an expression really
is an abbreviation for ‘an alternating-current voltage ” or for
“ alternating-current current.” Although one would not say the
latter, one often uses the abbreviation. In many radio circuits
there are both a.-c. and d.-c. branches and in some or all of them
both direct and alternating currents and voltages exist. It is
simpler to speak of an ‘“a.-c. voltage’ than of an alternating
e.m.f. and in the interest of simplicity this terminology has been
employed here whenever it is useful.
90. Effective value of alternating voltage or current.—Since
the voltage (or current) in an a.-c. system is rapidly changing direc-
tion, and since the needle and mechanism of an ordinary d.-e.
measuring instrument require appreciable time fora deflection, they
cannot follow the rapid changes of voltage or current and would
only wobble about the zero point of the meter even if they could fol-
06 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
low the fluctuations. We can, however, compare direct and
alternating currents by noting their respective heating effects.
An alternating current is said to be equal in value to a direct cur-
rent of so-many amperes when it produces the same heating
effect. This is known as the effective value of the alternating
current and is equal to the maximum values multiplied by .707 or
divided by V2.
Ty Rah ogall of ee Cay beg be Sora. He =
where I, = effective value of an alternating current;
J = maximum value.
The effective value is also known as the root mean square or
r.m.s. value for the following reasons. The heating effect of a
direct current is proportional to the square of the current. Then
if we take the instantaneous values of the current over a cycle of
alternating current, square them, get an average of these values,
extract the square root of this value, it will be equal to the direct-
current value that will produce an identical heating effect. The
value of current secured in this manner is .707 times the maximum
value. Since it is the “square root of the average or mean
squares ” of several current values, it is abbreviated to the “ root
mean square ”’ orr.m.s. Anr.m.s. voltage is one that will produce
a current whose heating effect is the same as a given direct current
as discussed above.
H
Liccem — E, et 107 Een => a
: V2
and
= Ber.
eee cor
Voltage or current is considered as effective unless otherwise
indicated or stated.
E
= Een. x V2
Example 2-6. What is the effective value of an alternating voltage whose
maximum value is 100 volts?
Een. = .707 X 100 = 70.7 volts
Problem 4-6. The effective value of an alternating current is 12 amperes.
What is the maximum value?
PHASE RELATIONS BETWEEN CURRENT AND VOLTAGE 97
Problem 5-6. The maximum value of an alternating voltage is 110 volts.
What it the effective value?
Problem 6-6. At the 45° phase the instantaneous value of an a.-c.
alternating current is 10 amperes. What is its effective value?
Problem 7-6. The effective value of an alternating current is 100 milli-
amperes. What is the instantaneous value at the 60° phase?
91. Phase relations between current and voltage.—Whenever
an a.-c. voltage forces a current through a resistance, the wave form
of the voltage and the current, the mathematical formula for them,
and the vector diagrams look alike. This is explained by the fact
that the current and'voltage start at the same instant, rise to a
maximum value at the same instant and carry on throughout their
respective cycles in perfect step, or in phase.
When an inductance or a capacity or any combination of these
quantities with each other or with resistance is in the circuit, other
phenomena take place differing entirely from what happens in a
d.-c. circuit. For example, when an a.-c. voltage forces a current
through an inductance, the current does not attain its maximum
value at the same instant as the voltage, but at a later time; when
the inductance is replaced by a capacity, the opposite is true, the
maximum value of the current takes place before the maximum
voltage is reached.
Fia. 74.—Current and voltage in phase.
92. Case I. Current and voltage in phase.—Figure 74 represents
the current and voltage in phase, i.e., in a resistive circuit. Since
98 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
the form of the voltage and current waves is exactly similar they
can be drawn on the same horizontal axis, or in the vector diagram
can be represented as in Fig.
75. They can be thought of
as two vectors, which may or
may not have the same length,
rotating at the same speed in
two different circles which
move forward in the same
direction at the same speed.
Under these conditions, the
end point of the vectors will
trace out identical curves.
ee ee
ps
Fra. 75.—Current and voltage in phase.
The maximum value of voltage greater ;
(or drawn to different scale) than the In such cases Ohm’s law
maximum value of the current. I= HKH/RF tells us the rela-
tions between current, volt-
age, and resistance, just as it does in a d.-c. circuit.
Example 3-6. Suppose a lamp of 55 ohms is placed across a 110-volt
60-cycle a.-c. line. What current will flow through it at the 30° phase?
We must first find the maximum value of the voltage.
E = Een, X V2
= 110 X 1.41 = 155 volts.
Since there is no phase effect in the circuit due to the resistance, the current is
given by Ohm’s law
l= HR
= —— = 2.82 amperes.
This is the maximum current. The current at any phase may be found by
a
I sin 6
2.82 sin 30°
= 2.82 < .5 = 1.41 amperes.
ll
Problem 8-6. A resistance of 10 ohms is in a cireuit with an alternating
voltage of 20 volts maximum. At what phases will the current through the
resistance be 1 ampere?
PHASE RELATIONS BETWEEN CURRENT AND VOLTAGE 99
Problem 9-6. A certain alternating current has the same heating effect
as a direct current of 8 amperes. It flows through a resistance of 25 ohms.
What is the effective and maximum voltage? At what phases will the instan-
taneous value of the voltage be equal to one-half the maximum value?
93. Case II. Current lagging behind the voltage.—Let us con-
sider the case where the current does not attain its maximum
value at the same instant that the maximum voltage is reached, as
is illustrated in Fig. 76. It will be noted that the current curve
does not start until 67.5° of the voltage curve has been completed,
and therefore that the maximum value of the current is said to lag
behind the voltage maximum 67.5°. The current and voltage in
<—67.5—>! Ke—-67.5—>) | | be-67.5—>1
<—67.5—>
|
Fic. 76.—Current and voltage in an inductive circuit where the current lags
behind the voltage.
such a case may be thought of as two vectors, or arms, moving in
two circles one of which, the current circle, does not start until the
other or voltage circle has completed 67.5° of its total movement
of 360°.
The formulas for Case II where the current is lagging are
E sin 0,
4 = Isin (6 — 9),
where § = phase of the voltage in degrees;
¢ = difference in phase between £ and J or the angle of lag.
(The angle ¢ is called “ Phi.”’)
é
100 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
Example 4-6. The current lags behind the voltage by 60°. T he maximum
value of the current is 40 amperes. What is the instantaneous value of the
current at the 75° phase?
Solution. Lay off on graph paper the voltage at the 75° phase and
60° behind it the current which has a maximum value of 40 amperes. The
vertical component then is equal to the instantaneous value of the current at
this phase.
The problem may also be solved by the mathematical formula
4 = I sin (75° — 60°)
= 40 sin (75° — 60°)
= 40 sin 15° = 40 X .26 = 10.4.
Problem 10-6. If the maximum voltage is 110 what is the instantaneous
voltage at the 110° phase? At the 90° phase? At 45°?
Problem 11-6. In an inductive circuit there is a phase difference of 25°.
When the voltage is a maximum, the instantaneous value of the current is
10 amperes. What is the maximum value of the current?
The cause of lagging current is inductance, which tends to make
the maximum of current take place later than the maximum of
voltage. If a circuit is purely inductive (the resistance is negligible)
the difference between these maxima is 90°. If there is appreciable
resistance the difference is less than 90°.
94. Inductive reactance.—The opposition to the flow of cur-
rent which inductance imposes on a circuit is called the inductive
reactance and is measured in ohms just as resistance is. Its
abbreviation is X;. In any circuit in which there is only resistance,
the expression which connects voltage and current is the familiar
Ohm’s law,
oltage E
current = posses oe (a=
resistance ifs
Similarly, the expression when inductance is in an a.-c. circuit
is
voltage E
current = ———— or =,
reactance AL
and if the voltage is the maximum value the current will be the
maximum value; if the voltage is the effective or r.m.s. value, the
INDUCTIVE REACTANCE 101
current will be the effective value; if the voltage is the instan-
taneous value the current will be the instantaneous value.
Inductive reactance is numerically equal to
Xz (ohms) = 6.28 Xf X DL =-ol
where f = frequency in cycles per second;
L = inductance in henries;
Greek letter omega = 6.28 x f.
@
Example 5-6. In an a.-c. circuit the following data are given: Hy = 110
volts; inductive reactance, X; = 20 ohms. Find the maximum and effective
current and the instantaneous current at the 150° phase?
Solution. The effective current is found from
Ey
Vf =
f re
110
= — = 5.5 amperes.
20
Imax = Iy X 1.41 = 7.8 amperes.
The vector diagram in Fig. 77 shows the instantaneous current to be
4 = Isin (6 — 90°)
= 7.8 sin (150° — 90°)
= 7.8 sin 60°
= 7.8 X 0.866
=: 6.75 amperes.
Problem 12-6. In the above example
find the instantaneous voltage when the
instantaneous current is 6 amperes. At
what phase is this?
Problem 13-6. What inductive reac-
tance is needed to keep the maximum pV, 77 Current eon hind
current down to 75 amperes in a 110-volt the yaltace by 90°.
circuit?
95. Case III. Current leads the voltage.—In this case the max-
imum value of the current is reached before the corresponding
maximum voltage is reached. The voltage lags behind the cur-
102 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
rent, or as it is usually stated, the current leads the voltage. A
vector diagram for such a case is shown in Fig. 78.
In this case the instantaneous values of voltage and current are:
e = E sin 60°
i = Isin80° or i= Jsin (60° + 20°)
The formulas for Case III when
the current leads the voltage are
e = Esin8,
I sin (6 + 4),
Fra. 78.—Current leading the Where ¢ is phase difference between
voltage by the angle ¢ at the 60° E and I or the angle of lead.
phase. The angle of lead is 20°.
a
The current in such an equation
will be the maximum current if the voltage is maximum, effective
if the voltage is effective, etc.
Example 6-6. The effective current in an a.-c. circuit is 70 amperes. The
angle of lead is 30°. What is the instantaneous current when the voltage is
at the 10° phase?
The maximum value of the current is found from
Imax= Jen X 1.41
= 70 X 1.41
98.7 amperes.
Il
By the equation
27 = T/1sin (6+ 4)
98.7 sin (10° + 30°)
= 98.7 X .643
= 64.5 amperes.
Problem 14-6. The instantaneous value of current in a certain capacitive
circuit is 8 amperes. The instantaneous value of the voltage is 25. The
maximum values of the current and voltage are 15 amperes and 80 volts
respectively. What is the angle of lead between them?
96. Capacity reactance.—The opposition which a condenser
offers to the flow of current in an a.-c. circuit is called its capacitive
CAPACITY REACTANCE 103
reactance and is measured in ohms just as resistance and inductive
reactance are. The equation
voltage E
current = Soest Sane or J =—.,
capacitive reactance Xe
is similar in form to Ohm’s law and the equation for current in
an inductive circuit.
Capacitive reactance is equal numerically to
X-(ohms) = : = :
6.28 xXfXC ow
where f = frequency in cycles per second;
C = capacity in farads;
w = Greek letter omega = 6.28 &X f.
Current leads the voltage in a capacitive circuit because capac-
ity tends to prevent any changes in voltage and so the maximum
of current in a purely capacity circuit takes place 90° ahead of the
maximum of voltage. If there is a appreciable resistance in the
circuit the difference is less than 90°; thus resistance tends to
bring the current and voltage in phase.
Example 7-6.—If a condenser which has a capacity reactance of 5 ohms,
is in an a.-c. circuit the instantaneous value of the voltage at the 20° phase
is 48 volts. What is the maximum current through the condenser? What is
the instantaneous current through it at the 20° phase?
Solution. é= EL sin 205
48 = FE X .342
E = 140 volts
E
(2 =
Xe
140
= a = 28 amperes
tz = I sin (20° + 90°)
4 = 28 sin 110°
28 cos 20°
= 28 X .940
26.32 amperes.
ll
104 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
Problem 15-6.—If a condenser in an a.-c. circuit whose voltage is 110
passes a current of 3 amperes what is the reactance of the condenser in ohms?
Problem 16-6.—Condensers are usually rated at the maximum voltage at
which they can be operated with safety. What should be the rating of a con-
denser to be used in a 220-volt a.-c. circuit?
Problem 17-6.—An a.-c. circuit has a voltage of 115, and a current of 4.5
amperes is flowing. It has a condenser in it. What is the reactance of this
condenser? What is the instantaneous value of the current when the voltage
is 80 volts? At what phase is this?
97. Comparison of inductive and capacitive reactances.—Coils
and condensers have opposite effects upon an alternating current.
The reactance of aninductance increases with increase of frequency;
a condenser has less reactance as the frequency increases.
A coil which will pass considerable current at 60 cycles may
pass practically none at one million cycles. Where it is desired to
pass a low-frequency current and to prevent the passage of a high-
frequency current, an inductance in series with the circuit may be
used. Note that the coil—called a choke coil—is in series with the
circuit, where a by-pass condenser would be across the circuit. If
even greater discrimination against a high-frequency current is
desired, a combination of condenser and choke is used. The choke
is in series, the condenser is shunted across the circuit.
Example 8-6.—Assume an a.-c. circuit composed of an inductance of one
millihenry. What current will flow if # is 100 volts and the frequency is
600 cycles?
XG, HOPS SP YS Ib
= 6.28 X 1 X 107° 600
= 3.8 ohms
i a ey 26.3
¥, 580 amperes.
When it is desired to exclude current of a given frequency from a circuit,
a shunt capacity is placed across the circuit. High frequencies will go through
_this by-pass condenser whereas the lower frequencies will not be so shunted
and will go on through the rest of the circuit. A frequent use is where an
R.F. choke passes direct current but prevents the flow of alternating current
whereas a condenser passes a.c. but stops d.c. A large condenser is placed
across batteries so that a.-c. currents will bave an easy path around them.
MEASUREMENTS OF CAPACITIES 105
Example.— What is the reactance of a 500-mmfd. condenser tc radio waves
of a frequency of 600 kilocycles?
1
EX eee
6.28 X 600 103 x 500 x 10-12
iA 10°
~ 6.28 X 600 x 500
Problem 18-6.—What would the current be if the frequency were 6000
cycles? 60 cycles? If the inductance were one henry? One microhenry?
Assume E = 100.
Problem 19-6.—Calculate the reactance of one henry, one millihenry, one
microhenry at the following frequencies: 100 cycles, 1000 cycles, 1,000,000
cycles.
Problem 20-6.—What reactance is needed to keep the current into an
electric iron down to 5 amperes when it is placed across a 110-volt circuit
(assuming the iron has no resistance)?
Problem 21-6.—The inductance in an a.-c. circuit is .4 henry. At what
frequency will the current be 3 amperes if the voltage is 110?
Problem 22—6.—Calculate the reactance of a 1-mfd. 0.001-mfd., 50-mmfd.-
condenser at 60 cycles, 60,000 cycles, 600 ke., 15,000 ke.
Problem 23-6.—The capacity in an a.-c. circuit is 1.0 mfd. At what fre-
quency will the current through it be 415 milliamperes if the voltage is 110?
98. Measurements of capacities—Condensers may be meas-
ured for capacity by comparing them with a known capacity by
means of a Wheatstone bridge. In this case the unknown capacity
may be calculated from
which differs from the formula when resistances or inductance are
measured because of the fact that the reactance of a condenser
decreases the larger it is. The formula when inductances are
compared is:
The capacity of condensers from 0.01 to 10or more microfarads
may be measured by noting the current through them with a known
106 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
voltage across them. The condenser is first tested for open or short
as indicated in Section 81. Then, in series with a milliammeter, the
condenser is plugged into a light socket (a.-c. of course). Then the
voltmeter may be put across the condenser, in case the voltage of
the line is not known. The capacity is
Ima. X 1000
CG mfd. a3 6.28 XfX FE
99. Combinations of resistance with capacity or inductance.—
Coils and condensers are never pure reactances. They always have
some resistance in them, although in most radio apparatus the
resistance is negligible compared to the reactance. Since a reac-
tance as well as a resistance impedes the flow of current, we must
combine them to determine what current will flow through a piece
of apparatus under a certain voltage and at a certain frequency.
Because an inductance has a different effect upon an a.-c. cur-
rent than a resistance, and different from capacity, we cannot
merely add their reactances in ohms to determine the resultant
effect upon the circuit. They must be added vectorially, not
algebraically.
100. Impedance.—Combinations of resistance and reactance
are called impedances. The value in ohms may be found as fol-
lows: Two factors whose effect is at right angles to each other
may be combined and the resultant secured by the formula found
in plane geometry which states that ‘‘ the square on the hypot-
enuse of a right triangle is eaual to the sum of the squares on the
other two sides.”” Thus
72 = R? + XG oe
and if R = 3 ohms and XY = 4 ohms,
whence
Z=V25 = 5.
This is called getting the vectorial sum. The algebraic sum, ob-
tained by simple addition—as when two resistances are combined—
would be, in this case, 7 ohms whereas the vectorial sum is 5 ohms.
GENERAL EXPRESSIONS FOR IMPEDANCE 107
The resultant of combining a resistance and a reactance can be
found graphically. Lay off on a horizontal line a number of units
corresponding to the number of ohms resistance. Then if the
reactance is inductive, erect a perpendicular and lay off on it a
number of units equal to the number of inductive reactance ohms.
The length of line connecting the extremities of these two lines is
the resultant impedance in ohms.
Because capacitive reactance has an opposite effect to that of
inductive reactance, the line representing it should be pointed
downward. Graph paper is of great aid in solving a.-c. problems
in this manner.
Example 9-6.—An alternating current of 8 amperes maximum flows
through a coil whose inductance is 0.043 henry and whose resistance is 5 ohms.
What voltage is required if the frequency is 60 cycles?
The current in such a circuit is
[= B/Z
Z=VR?+ XxX?
and X =2xf X L=6.28 X 60 X .048 = 16.25 ohms
Z = V5? + 16.25? = V289 = 17 ohms
whence EEN Gs — L36rvolts:
Example 10-6.—What is the impedance in a circuit in which there is a con-
denser of 1.66 mfd. and a resistance of 800 ohms? The frequency is 60 cycles.
TaN he xe
1
X, = 2 fC
Leer aii RA, a
6.28 x 1.66 X 60
Z = V 8002 + 15902
V (64 X 104) + (256 X 104) (approx.)
102V'320
= 1790 ohms.
= 1590 ohms
101. General expressions for impedance.—If an a.-c. circuit
is composed of resistance and both inductive and capacitive reac-
tance the impedance is figured as follows. Since the inductive
ractanea, That & a capacity rmactaace bs a Begative .
of spanany ahas: an Raleetive reactance A a PARTE Feaetane
SS SQmmany chag Refs they are combed WRA the raaatance
Yeetorialy, thet akehak aa ik
Z=aeVRE+OG — VOX
aad after the capacity reactance hag”
Rue deem comdeted wRA the eetive
Tada UR BR actaaly @derncted
because the seas af the dee Ran
tances are Gidrend) the Qa af the
Bapedace Demees as bala,
be will be noted that the Sea hetee
T XS & the adeve equation & paative. |
Tas B&B abaus De aS Mee tre
t BERATING QUARTRIES MARI together
Tt
T|
.
{or Bs negative quantity & gaara)
reat lh a PoRtive gaat, BL Re)
Smale, the actual walke ef the cage
RY ractanee a ahas WAS greater
than the tadeetive reactance 2 cha
Ris. TA— Vector dagramef a the edtetive Raa Re the akealt
Gc ements OU BS meeave Bat EE Weald Be
@edms) and nduetie Rag PWUW
Race UO chase.
Ramses weword at an angle of 20° tron the nechtancte aad Xi pointing dune
ward af am ange of WO them the resktanen. ‘The wral eiRet ef de mac
Renees SIO — T or s pusitive 3 ohms witch pels Upeand
SERIES A.-C. CIRCUITS 109
If, however, the values of X,, and X; are interchanged, so that
the resultant of adding the reactances is a negative 3 ohms which
point downward, then
VP pax
= VEER aS,
In Case 2, Z= VR? + (X, — X)2
~ VP4 0-10
es:
a V/ 42 2932
= 5.
In Case I, Z
I
Problem 24-6.—An antenna (Fig. 80) may be considered as an inductance,
£, and a condenser in series.
If the voltage in Fig. 80 is
100 microvolts f = 1000 ke.,
what is the current through oe
the coil L, in series with L,? een
(There is no mutual induc- iB,
roar bao
{
I
!
i
tance between L, and L;.) —>L,=100uk
Problem 25-6.—What is
the effective reactance in a
circuit which has 45 ohms
inductive reactance, 70 ohms
capacitive reactance and 20
ohms resistance? What is
the impedance? What cur-
rent would flow if the voltage Fic. 80.—An antenna and its equivalent.
were 110 effective?
Problem 26-6.—What would be the values of capacity and inductance if
the frequency were 500 cycles?
Problem 27-6.—What voltage is required to force one milliampere through
the following circuit: Resistance 8 ohms, inductance 300 microhenrys,
capacity 500 mmfd., frequency
Ls —>Ls=50uk
il
R
eet ogcgo = WWVWy-——— 750 ke.?
7= ./R+ 2 w? 102.—Series a.-c. cir-
cuits—In Fig. 81 is an
inductance in series with
a resistance. The current flowing in the circuit may be found
Fig. 81.—A series circuit: w = 6.28 & f.
110 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
by dividing the voltage across the circuit by the impedance of
the circuit. That is: J = H/Z, in which
Z= VR? + X?,
which is quite different in numerical value from K+ X. For
example, if R = 3 and X = 4, the vector sum Z = 5 where as
the arithmetical sum = 7.
As in a d.-e. circuit, the voltage across an impedance, a reac-
tance, or a resistance is equal to that impedance, reactance or
resistance in ohms times the current in amperes.
Voltage across a resistance Be= DT XxOR
Voltage across an inductance Hy, =I X Xx
Voltage across a capacity Eo =I X Xe
Voltage across L and C Er+c= I (Xz — Xe)
The voltage across two resistances or reactances is the algebraic
sum of the individual voltages, remembering that a capacity reac-
tance has a negative sign and that the voltage across it is negative
with respect to that across an inductance. The voltage across two
impedances, however, must be determined by adding the individual
voltages vectorially. This is because the impedance is a vector sum
of a resistance and reactance.
Let us take a typical example. The current in the circuit of
Fig. 81 is
E
1 pe
or E=IVR?+X2
BE? = [? (R2 + X2)
or H? = [?R2 + [2X2
= Hp? + Ex?
whence E=VE,24+ Ex?
Therefore the resultant voltage across a resistance and a reactance
is the vector sum of the individual voltages.
PHASE IN SERIES CIRCUIT ig
Example 12-6.—If E = 15, R =3,X =4
15 15 15
I= — —=>
Vile? SEI 2a Ghiig ons
= 3 amperes
Hr= IR = 3 <3 = 9 volts
Ex = IX =3 <4 = 12 volts
E = VEp? + By? = V81 4144 = V285 = 15
Experiment 1-6.—To measure the capacity of a condenser. Place a con-
denser of about 10-mfd. capacity in series with an a.-c. milliammeter and
measure the current through it when placed across a 110-volt, 60-cycle line.
1
Then 1 S112, 2 1 <<
6.28 Xf XC
or (Gee ee,
E X 6.28 xf
where FH = line voltage
I
or Canidae
m AS x .024
when EF = 110;
f = 60;
I = milliamperes.
103. Phase in series circuit.—In a resistive circuit, the voltage
and current are in phase, their maximum values occurring at the
same instant. If the circuit is purely capacitive, or inductive,
there is a 90° phase difference between >
current and voltage.
If, instead of a pure reactive cir-
cuit, we have some resistance, the
angle of lead or lag (Sections 93, 95)
is not 90° but is some value less than ; ZA) B
this. To determine the phase angle, ay eine ence
or the difference in phase, let us draw jap Ponrosonts tthe ecto
the vector diagram of the voltages as voltage across an inductance
in Fig. 82. The voltage across the re- _and a resistance in series.
sistance, IR, is the horizontal line, and
at an angle of 90° with it is the voltage across the reactance, /X.
The diagonal of the parallelogram represents the resultant voltage
across the impedance, JZ.
112 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
Because a resistance in an a.-c. circuit produces no phase
difference between the current and voltage the current through
and the voltage across a resistance are said to be in phase. Their
directions may be represented along the same line, that is, along
the horizontal line representing the voltage across the resistance
IR. The direction of the diagonal represents the direction of the
voltage across the entire circuit. Since, then, the direction of the
TR line represents the direction of the current, J, and the direction
of the IZ line represents the direction of the voltage, H, the angle
between these lines represents the angular difference in phase
between the voltage and current. The angle 6 then is equal to the
angle of phase difference between the voltage across the combina-
tion and the current through it. Thus, in Fig. 82 BD + AB is
the tangent of the angle @, or
le
AB aes
and since BD = AC = IX (or the voltage across X) and AB =
IR (or the voltage across F),
Emel
Halk eR
= tan 0.
Knowing the reactance and the resistance in ohms the tangent
of the angle may be determined, and the angle itself looked up in
a table. When the tangent of an angle is known but not the angle,
=
the expression is written 6 = tan—! = and is read ‘‘@ (theta) the
the angle whose tangent is XY over R.”’
The effect of a resistance in series with a reactance is to decrease
the angle of phase difference between the current and the voltage
or to bring them more nearly in phase. In a pure reactance circuit,
the angle is 90°; when resistance is added this angle decreases.
In a pure resistance circuit, there is no angle, the current and voit-
age are in phase; they reach their maximum values at the same
instant.
CHARACTERISTICS OF A SERIES CIRCUIT 113
Example 13-6.—In an a.-c. circuit there is a resistance of 10 ohms and an
inductive reactance of 8 ohms. <A current of 8 amperes is flowing. What
voltage exists across each part of the circuit and across the entire circuit?
What is the phase difference between the current and the voltage?
Voltage across R = IR = 8 X 10 = 80 volts
Voltage across X = JX =8X8 = 64 volts
Draw the vector diagram to scale as in Fig. 83; then the diagonal JZ =
102.5 volts. (Note that the algebraic sum of the voltages across R and X is
144 volts.) The tangent of the angle of phase difference is equal to X + Ror
tan @ = X/R = 8/10 or 8
Oi Tanwe.S
or Ga—toome
Tan g-et=8
6 = 38° 40'
Fic. 83.—A vector diagram of problem 13.
104. Characteristics of a series circuit—1. The voltage across
a series combination of resistance and reactance is the vector sum
of the voltages across the separate units. ey
2. The combined resistance of several resistances in series 1s
the algebraic sum of the individual resistances.
3 The combined reactance of several reactances, whether
inductive or capacitive, is the algebraic sum of the individual
reactances.
114. PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
4. The impedance, or combined effect of a resistance and reac-
tance, is the vector sum of their individual values.
5. The combined impedance of several separate impedances is
the vector sum of the individual impedances.
Example.—Suppose we combine two resistances, reactances, etc., of 3 and
4 ohms respectively. The table below shows the resultant values.
Combination Sum Resultant
1. R=3; X = V9+16 =Z 5
9.R=3; R=4 Sena =F 7
oy bee a hese 4253 x 1
3. X_, = 3; X= 4 3-4 = 3K —1
B 2G, SBR Ms, = 443 = 2K if
By Dies = GB OG Se —4—3 = —7
4,.R=3;X =4 N/ Oe 16 ae,
(1) R=3,Xr =4 X-
(OD) is = ay 2G, 8) 2G
(3) Rh =3, Xr =3 Xc
v
ll
ll
MAU Trae be ay) ae
SV ATE UE SAG a5 52
II
5
3 V94+ (4-3)? =Z 3.16
3
3
Note in (1) and (2) above that the resultant is the same although the
conditions are different. This is due to the fact that a negative number when
squared, or multiplied by itself, becomes a positive number. In other words,
a negative reactance and a resistance always produce a positive impedance.
105. Resonance.—In (3) above, a very important phenomenon
is illustrated. When the capacity reactance of a series circuit
equals the inductive reactance, their respective effects cancel out
and the resultant impedance is the resistance in ohms alone. To a
circuit possessing inductive reactance one can add a certain
amount of capacity and thereby reduce the impedance of the
circuit (at some particular frequency) to the value of the ohmic
resistance. This is the phenomenon underlying all tuning in
radio circuits; it is known as resonance.
Problem 28-6.—Two resistances, one of 8 ohms and the other of 24 ohms,
are in a 60-cycle 110-volt circuit. What current flows? What is the current
if the frequency is increased to 500 cycles?
PARALLEL CIRCUITS 115
Problem 29-6.—What current exists in the above-named circuit if the
second resistance is replaced by an inductive reactance of 24 ohms? If the
frequency is 60 cycles what inductance will be in the circuit?
Problem 30-6.—In a circuit with 550 volts across the terminals are the
following pieces of apparatus: A coil with 15 ohms reactance, a condenser
with 7 ohms reactance, two resistances of 10 and 5 ohms. What current
flows? What is the voltage across each part? What is the phase relation
between voltage and current?
Problem 31-6.—In an a.-c. circuit appear a voltage across a resistance of
34 volts and a voltage across a capacity of 66 volts. What is the voltage
across the combination?
Problem 32-6.—Two condensers are in series with two inductances and a
resistance. The condensers have reactances of 8 and 10 ohms, the inductances
20 and 6 ohms, and the resistance is 4 ohms. What current flows, what
voltage appears across each component and what is the phase between current
and voltage? Assume # = 110.
Problem 33-6.—What is the phase difference in the following cases: (a)
pure resistance circuit; (b) pure inductive circuit; (c) pure capacity circuit;
(d) 100 ohms resistance and 100 ohms inductive reactance; (e) 100 ohms
resistance and 50 ohms inductive reactance; (f) 100 ohms resistance and
100 ohms capacity reactance; (g) 100 ohms resistance and 50 ohms capacity
reactance; (h) 100 ohms inductive reactance and 50 ohms resistance and
25 ohms capacity reactance; (i) 100 ohms each inductive and capacitive reac-
tance and 100 ohms resistance?
Problem 34-6.—In a series circuit there are 45 ohms inductive reactance
and 20 ohms resistance. It is desired to increase the phase angle between
the current and voltage to 85°. How can
this be done? How can the phase angle be I, L
decreased to 30°?
106. Parallel circuits—In a cir-
cuit like that of Fig. 84, in which
several reactances or combinations
of reactance and resistance may be
connected in parallel, the following
rules hold: ay h Fig. 84.—In a parallel tcircuit
The voltage across eac Tanch the current J may be very small
equals the voltage across the com- compared to Iz or Ic.
bination.
The current through the combination is the vector sum of the
currents through each branch.
116 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
The impedance offered to the flow of current by the combina-
tion is the voltage divided by the total current.
Thus in Fig. 84 the current through the entire combination
may be found as follows, assuming H = 120; Xc = 8; Xi = 5;
Riss3:
Ic = ue = es = 15 amperes through the condenser
Xe 8
Ir = ‘ = = = 40 amperes through the resistance
Ilh= a = 2 = 24 amperes through the inductance
L
fe AV Tx + Ud, — Ic)? = V 1681 = 41 amperes
Tt ea = 2.92 ohms.
4]
107. Phase in parallel circuit—The phase angle between the
current and the voltage in a parallel circuit may be obtained from
the expression
In Ic
Tp
I,—TI
or 0 = tant ( 7 ‘)
R
tang
108. Impedance of parallel circuit.—Since the impedance is the
ratio between the voltage across the circuit and the current through
it, in order to find the impedance of several branches in parallel we
must know the voltage and the current. Often we would like to
know the impedance without knowing either the voltage or the
current. The procedure then is to assume a voltage, to find the
currents that would flow, divide the voltage by the current and
get therefrom the impedance, or
Z=
No
POWER IN A.-C. CIRCUITS 117
Example 14-6.—What is the impedance of 630 ohms capacity reactance
shunted by 100 ohms of resistance?
Assume a voltage of 100.
Ic = 100/630 = .159 ampere
Ip = 100/100 = 1.0 ampere
Total current J = V Ic? + Ip?
= V 1592 + 12 = V'1.025
= 1.015 amperes
Fic. 85.—All the power in a resistive circuit is used up.
109. Power in a.-c. circuits.—In a d.-c. circuit the power is the
product of the voltage across the circuit and the current through
it. Thus, if one ampere is fed into a device under a pressure of
100 volts, the power used is 100 watts.
In a.-c. circuits the voltage and current are not always in phase.
In fact in many circuits there is a decided difference in phase
between the current and voltage. What is the power?
The power at any instant is the product of the instantaneous
current and the instantaneous voltage. Thus in Fig. 85, where the
118 PROPERTIES OF ALTERNATING-CURRENT CIRCUITS
voltage and the current are in phase, as in a resistance circuit, the
height of the voltage line e above the horizontal, or time, axis
multiplied by the height of the current line 7 above this axis gives
the instantaneous power.
This is plotted in the
yp curve P.
When, however, the
current and voltage are
g not in phase, as in an
YY inductive (Fig. 86) or
J capacitive (Fig. 87) cir-
cuit, a different looking
Fic. 86.—In an inductive circuit some power Curve results although
is consumed and some (the shaded areas the instantaneous power
below the line) is returned. is still the product of the
instantaneous values of
current and voltage. The part of the power curve marked B is
interesting. It is the result of multiplying a positive current by a
negative instantaneous value of voltage. The product is negative;
so the power at that instant represented by this small loop must
be considered as nega-
tive power. What does
this mean?
Power consumed in a
circuit is considered as
positive power. Nega-
tive power is power that
is returned to the gen-
erator from the line.
Power is only returned
to the generator when
there is reactance in the circuit. A pure resistance circuit con-
sumes the entire amount of the power fed it by the generator;
a reactive circuit returns some of it to the generator.
The effective power in a resistance circuit is the product of the
effective volts and the effective amperes. In a reactive circuit,
however, the effective power is reduced by the power returned to
Ve
Fig. 87.—Power in a capacitive circuit.
POWER IN A.-C. CIRCUITS 119
the generator, so that the product of effective volts times effective
amperes does not give the true measure of the power consumed in
the circuit. The true power is given by
P = Eyl, cos 8,
where @ is the angle between the voltage and the current.
The product of the volts and the amperes is called the apparent
power. Since this apparent power must be multiplied by cos 6,
this factor is called the power factor of the circuit. When the cur-
rent and voltage are in phase, that is, in a resistance or resonant
circuit, the power factor, cos 6, is equal to 1.0, and the circuit is
said to have unity power factor.
CHAPTER VII
RESONANCE
Tux most important circuits in radio are those in which either
series or parallel resonance occurs. In transmitting and receiving
systems resonance is used to built up large voltages and currents at
certain desired frequencies and to discriminate against undesired
signal frequencies by keeping their voltages and currents low.
When one tunes a radio receiver, he actually adjusts the a.-c. cir-
cuits within the receiver so that a condition of resonance occurs.
Everyone who has operated a receiver has, in tuning it, performed
one of the most interesting experiments in all a.-c. theory and
practice. It is necessary that we look into the phenomenon of
resonance very closely.
110. Series resonant circuit Although a general idea may be
obtained of what takes place in a resonant circuit when a radio
/ . receiver is tuned, a much more
exact idea may be had as a re-
sult of a laboratory experiment.
Experiment 1-7.—Connect, as in
Fig. 88, a coil of about 200 micro-
henries inductance, a variable con-
denser of maximum capacity of
1000 mmfd., a resistance of about
= 10 ohms, and a current-indicating
device such as a current squared
meter or a thermocouple and meter.
Couple the circuit loosely to a radio-
frequency generator and (a) adjust
the condenser C so that resonance
is obtained and then (b) adjust the
frequency of the oscillator while the tuning condenser C of the external
circuit is held constant. Plot the current in the circuit against condenser
120
Fig. 88.—When L is coupled loosely to
a generator and C is varied, the current
indicated at J will go through a maxi-
mum like that in Fig. 90.
SERIES RESONANT CIRCUIT 12
degrees or capacity and then against frequency. Change the value of R
and repeat.
In either case the voltage across the condenser and across the coil and the
phase between current and voltage should be calculated and plotted.
Nors.—If the experimenter possesses a short-wave transmitter which is
equipped with an antenna meter he can carry out the same experiment by
noting the antenna current as the antenna series condenser is adjusted, or as
the frequency of the closed circuit is adjusted below and above resonance with
lamp lights up. Adding the capacity has
brought the circuit to resonance so that 30 |
the only hindrance to the flow of current |
was the lamp and the resistance. Adding
something to the circuit actually made
more current flow.
the antenna. Such a curve is shown in 55
Fig. 89.
Experiment 2-7.—Connect in series Pa
with a lamp an inductance of several
henrys. Add sufficient resistance so that ;
the lamp does not light when placed < * i
across a 110-volt 60-cycle line. Then = —|
put a condenser (of the filter type) in s 40
series with the resistance, the line, and 5
the lamp. Add other capacity until the £ 35 | |
<
25
Ze 25 > 60
The curve in Fig. 90 shows what :
happens as the voltage across a
series circuit is kept constant but
the frequency is increased. At first
the current increases slowly, then as the resonant frequency,
356 ke., is approached the current increases very abruptly and after
passing through a sharp maximum at 356 ke. falls very rapidly at
first and then more slowly. The voltages across coil and condenser
go through similar changes. The phase between current and volt-
age changes also, being a negative angle (current leading voltage)
below resonance, being zero at resonance (current and voltage in
phase), and becoming a positive angle above resonance (current
lagging behind voltage).
At zero frequency, that is at direct-current, the current in
such a circuit would be zero because the condenser will not per-
mit d.-c. current to pass. At very low frequencies, the reactance
Fig. 89.—How the antenna cur-
rent of a radio station varies
as the series condenser is varied.
typ]
“129 RESONANCE
of the condenser is very high, so that little current will flow. At
very high frequencies the reactance of the coil becomes very great
and therefore little current will flow. At intermediate frequencies
more current flows.
When a series circuit is resonant, the current and voltage are in
phase, the current is a maximum, the impedance is a minimum, the
voltages across the condenser and the inductance are equal and
1.0
9 “hae “le
‘ al
F |
al 8000
Cycles
pet.6
a
oe eal oy
A
|
290 300 310 320 330 340 350 360 370 380 390 400 410 420
Ff KC
Fria. 90.—The resonance curve of a circuit like that of Fig. 89.
opposite in sign and greater in value than the voltage across the
combination.
In Fig. 90 note that from 340 to 356 ke., a change of 1.047 times,
the current changes from 0.19 amperes to 1.0 amperes, a change of
5.2 times. The voltages across the condenser and inductance
become much greater, at resonance, then the voltage impressed
upon the circuit. This voltage may become so high that the
condenser will be punctured. The voltage across the coil or the
condenser at resonance is equal numerically to the voltage across
CHARACTERISTICS OF SERIES RESONANT CIRCUIT 123
the entire circuit multiplied by the factor X,/R or Xc/R which
are equal to Lw/R or 1/CoR.
A curve showing how the current in the circuit changes as the
variable factors are changed, that is, a graph of J against the
capacity, the inductance or the frequency, is called a resonance
curve and is symmetrical about the resonant frequency if the cir-
cuit is adjusted by changing the inductance, and is dissymmetrical
when the capacity or the frequency is the variable factor.
111. Characteristics of series resonant circuit—Below the
resonant frequency the reactance is mainly capacitive; above this
frequency the circuit is mainly inductive. That is: the capacitive
reactance is the main deterrent to the flow of current below reso-
nance; above resonance the inductance offers the greatest opposi-
tion to the flow of current. For a narrow band of frequencies in
the neighborhood of 356,000 cycles the total impedance of the
circuit is less than 100 ohms. Far from the resonant frequency the
reactance is much greater and very little current will flow.
Below resonance, where the capacity reactance predominates,
the current leads the voltage; at resonance the current is in phase
with the voltage; above
resonance the current
lags behind the voltage.
At all frequencies the
voltage across the induc-
tance is 90° ahead of the
current and the voltage
across the condenser is
90° behind the current.
Between the two reac-
tive voltages, then, is a
180° phase difference.
That is, they are exactly
out of phase. nes TC- Fg. 91.—Vector diagram of a series circuit in
sultant may be found by which inductance predominates.
looking at Fig. 91 in
which H, and Ez are plotted as 180° out of phase and of unequal
magnitude. The resultant of combining them with the voltage
124 RESONANCE
across the resistance must be the voltage across the entire series
circuit, which is the vector sum. ‘Thus,
E = V Ex? + (£1 — Ec)?
At any frequency but that of resonance, one of the two reactive
voltages is greater than the other. At resonance, however, the two
voltages are equal in magnitude and opposite in phase so that the
resultant of combining the reactive voltages is zero. When added
vectorially to the 7R drop in the circuit, the resultant is the voltage
which is impressed by the external source.
The vector sum of the reactive and resistive voltages 1s equal to the
impressed voltage.
Example 1-7.—What are the voltages and phase relations in the circuit
of Fig. 88 at a frequency of 370 ke.?
I = .274 ampere
Xc = 430 ohms
Xz = 466 ohms
R = 10 ohms
Er=I1X R= 274 X 10 = 2.74 volts
Eco =I X Xe = .274 X 430 = 118 volts
Ey, =I X Xz, = .274 X 466 = 128 volts
Ertit= V 2.742 + 1282 = 128 volts (approx.)
128
or+_L = tan OS = tan»! 46.6 = 88:46°
Ert+c = V 2.742 + 118? = 118 volts (approx.)
18
2.74
gr+c = tan“? = tan7!43 = — 88.38°
Er+c = Ey — Ec = 9.6 volts = Ey
E = IVR? +X? = V Ep? + Ex? = V2.7 +962
= 10 volts
is, X,—Xc a 466 — 430
R 10 eae
ae
¢ = tan-1— =
R+L+C an R
= 74° 30’
RESISTANCE IN SERIES RESONANT CIRCUIT 125
At resonance the reactances are equal to each other and equal
L raz
to Bs Ley X, = Xe = ae
C
For example the reactances of the condenser and inductance
in the circuit of Fig. 88 may be found by
; IL
=X = Ne
Bes Oe 200 x 10-6
1000 X 10-12
= V2 x 106
=V2~x 103
= 447 X 103
= 447 ohms.
At resonance the inductive reactance and the capacitive reac-
tance in the equation for the impedance Z = +/R? + (Lw — 1/Cw)?
cancel out, that is, Lw — 1/Cw = 0 so that the resultant impe-
dance is the resistance alone,
Z = R (at resonance)
Problem 1-7.—An inductance of .3 mh., a condenser of .0001 mfd., and a
resistance of 5 ohms are in series. Across the ends of this circuit is an alter-
nator whose frequency is 900,000 cycles and whose voltage is 5. Calculate
the current flowing, the phase between the current and voltage, the voltages
across the coil and the condenser. What would the current be if the circuit
were resonant? What is the impedance of the circuit at 900 ke.?
112. Effect of resistance on series resonant circuit.—At reso-
nance the magnitude of the current in the circuit is controlled
solely by the resistance. Its effect is most important in any radio
circuits where resonance plays a prominent part. The curves in
Fig. 92 show the effect of adding various resistance to the circuit
of Fig. 88. The voltages across the condenser and the inductance,
too, depend upon the resistance of the circuit. They are greater
the smaller the resistance. This is due to the fact that the voltage
across these reactances is equal to the product of the reactance
126 RESONANCE
and the current. The latter, controlled entirely by the resistance
at resonance, in turn produces greater voltages across the reac-
tance when less resistance is in the circuit. If H is the voltage
1.0)
R=100
I Amperes
R=20w
a R=300
325 330 335 340 345 350 355 360 365 370 375
FRC
Fia. 92.—Effect of resistance on a resonance curve. Note that the current
far from resonance is not ¢hanged so much as the resonant current.
impressed on the whole circuit, the voltage across the condenser is
E =~ CoR and that across the inductance at resonance is H X ze :
113. Power into resonance circuit.—No power is dissipated in
heat in a pure inductance or capacity, but energy stored at one
instant in a magnetic or electrostatic field is turned back into the
circuit at another instant. Power is expended in the resistance of
RESONANT FREQUENCY OF THE CIRCUIT 127
a circuit, but since at high values of resistance the current is
small, the power in the circuit will be small. This power is equal,
as usual, to
Pees 120.
where Ff is the resistance of the circuit. In case of Fig. 88, where
the resistance is 10 ohms and the current at resonance 1 ampere, the
power is 10 watts. Since at resonance there is no reactance effec-
tive in the circuit the power fed into it by the generator is the
product of the current times the voltage, or 10 X 1 or 10 watts.
In other words, all the energy taken from the generator is used
up in heating the resistance. None is necessary to maintain the
magnetic and electrostatic fields of the coil and the condenser.
The energy in these fields is transferred from one to the other, the
sum at any one instant being equal to the sum at any other instant
so long as the energy dissipated in the resistance is supplied from
the outside.
Problem 2-7. What power is taken from the generator in Problem 1-7?
In actual circuits the resistance is not isolated as in our dem-
onstration problems. All coils have resistance; so do all con-
densers, although the resistance of modern variable capacities is
quite small. These resistances take power from the generator
and reduce the maximum height of the resonance curve.
114. The resonant frequency of the circuit—The condition for
series resonance —that the reactances of the circuit add up to
zero—is fulfilled when
Xr — Xe
or Xp — Xo =
1
or wh — me 0
1
or wl = oC
or w LC
128 RESONANCE
and since 6.28 X f = wo,
1
TGs N LC.
and since 6.28 is equal to the mathematical expression 2 7 we
arrive at the familiar expression for the resonant frequency of a
circuit as
il
ages IE
in which f = the frequency in cycles;
L = the inductance in henrys;
C = the capacity in farads;
aw = the Greek letter ‘‘ Pi”’ and is equal to 3.1416....
Example 2-7. To what frequency will a circuit tune which has an induc-
tance of 0.25 henry and capacity of 0.001 mfd.?
Let us write the above formula as
1 1
2— =
f 4x? LO 39:5 LC
f= :
~ 39.5 X .25 X.001 x 1078
_ 109
~ 39.5 X .25
0:
9.87
f = V101 x V/108
10.1 X 10° cycles
= ON Ih ees
Such an expression for the resonant frequency of a circuit shows
that the frequency depends upon the product of L and C, and not
apon either of them alone. If L is doubled, C can still be halved
and the natural frequency of the circuit will not be changed.
WAVELENGTH 129
115. Wavelength.—The relation between the frequency of a
circuit and the wavelength of transmissions to which it responds is
a simple one. The wavelength is equal to the speed at which the
electric waves travel divided by the frequency in cycles. This
speed is 186,000 (approximately) miles a second and if we want
the wavelength in miles we need only divide this quantity by the
frequency. Ordinarily, however, we express wavelengths in meters;
so it is necessary to use the velocity of transmission in meters.
This is 800 X 10° meters a second, and so
300 x 108 300 X 10%
wavelength in meters = ————- or ———_—_..
3 f in cycles f in kilocycles
or = 300 X 103 X 2rV LC
The customary symbol for wavelength in meters is the Greek
letter ‘‘Lambda’’; so the above expression may be written:
7 300 X 10°
= = 1.884 VLC
kilocycles :
where JL = microhenry = 10-°H
Ci=immidy=— 10412 /.
Example 3-7. What wavelength corresponds to 1000 kilocycles?
300 < 10%
1000
I
A meters
Ow 2
10 ae
300
Figure 93 is a graphical method of correlating L, C, and meters
of wavelength. Such a curve is called in England an “ abac.”’ A
table of “ LC ” products will be found inside of rear cover.
Problem 3-7. What inductance must be placed in series with a 2-mfd.
condenser to resonate at 60 cycles? If the voltage across the combination is
110 (effective) and the resistances in the coil ‘and condenser add up to 20 ohms,
what power is consumed in the circuit at resonance, what is the resonant cur-
rent, and what voltage then appears across condenser and inductance?
130 RESONANCE
Problem 4-7. A coil of 0.15 henry is in series with a condenser of 28.5 mfd.
and a resistance of 5.8 ohms. The voltage across the circuit is 22 volts, the
frequency is the resonant frequency of 80 cycles. What would the voltage
across the condenser be if the resistance were doubled? What power is being
wasted in heat at resonance? ¥.
Problem 5-7. A variable condenser has a range from maximum to mini-
mum capacity of 9 to 1, that is, from 0.0005 to 0.0000555 mfd. What wave-
length range will it cover with a given coil, that is, what is the ratio between
the longest and shortest wave to which it will tune the coil?
100 LC =28.2 22
A is in Meters—Wave Length
L is in ~h—Inductance
C is in Mmfd.— Capacity
Wave Length
\
ZA \\
- ZZ" Ei\
DEES se
Vea Alo
AA) I; \o
N
eae al \ i
So oO Oo oO f=) i=) oO o- Oo fo) Oo oO oO oO oOo i=) So onw| ito
= a o Oo t+ OS Nos ono
~ ISS 8H SMR NNN HAS eey ce SN HS
Capacity
Fia. 93.—Drawing a straight line through two points (L and \ for example)
and intersecting the third line gives the unknown quantity desired (C).
Problem 6-7. In Problem 1-7 what would happen to the voltage across
the condenser if the capacity were reduced to half, resonance being maintained
by other means which also keep the original current?
Problem 7-7. An antenna may be represented by an inductance of 50
microhenrys in series with 0.00025 mfd. capacity and 30 ohms resistance.
What is its resonant frequency? If a distant station transmitting on this
frequency produces a voltage of 1000 microvolts across the ends of this antenna
system, what current will flow? What will be the ratio of currents at the reso-
nant frequency to signals of equal voltage at the antenna but differing from the
PARALLEL RESONANCE 131
resonant frequency by 50 ke., say 50 ke. above resonance? What is the res-
onant wavelength of the antenna?
Problem 8-7. It is desired to make a receiving set able to receive signals
from 800-meter stations. At present the maximum wavelength that can be
received is 600 meters. The tuning condensers have a maximum capacity of
500 mmfd. What capacity must be added in parallel to the present tuning
condenser to enable the 800-meter signal to be heard?
Problem 9-7. What can be done to increase the current at 150 meters in
the antenna of Problem 7-7? Suppose another 50-microhenry coil is placed
in series with the antenna. What must be done to bring the circuit to reso-
nance with the 150-meter signals? At resonance, what voltage will appear
across the 50-microhenry inductance, if C = 0.000126 mfd. and R = 30 ohms?
Problem 10-7. What power is being lost in this antenna at resonance?
Problem 11-7. The primary of a good audio-frequency choke coil has
100 henrys inductance. In many circuits a condenser is placed across the
primary so that high radio frequencies will not have to pass through the
transformer. If this condenser has
a capacity of 0.001 mfd., what is C=4.0 mfd
the decrease in effective impedance of
the circuit to a frequency of 10,000
cycles? ~
Problem 12-7. A loud speaker is werd
often coupled to a power tube through Loud Speaker
a condenser as in Fig. 94. If the =1.0H
speaker has an inductance of one henry
and the condenser is a 4-mfd. unit, to
what frequency will the combination
become resonant?
Problem 13-7. Plot a curve of the reactance of the loud speaker in Prob-
lem 12-7 from 100 to 10,000 cycles.
Problem 14-7. In an amateur’s short-wave transmitting station a 100-
mmfd. condenser is in series with the antenna. What voltage has this con-
denser across it, if the wavelength is 30 meters and the antenna current is one
ampere?
Problem 15-7. In a similar transmitting circuit an amplifier is used to
boost the output of an oscillator before being fed into the antenna. The grid
of this amplifier’s tube requires a voltage of 80 volts at 40 meters (7500 ke.).
This voltage is to be obtained across an inductance of 4.5 microhenrys. How
much current must flow through the inductance? What capacity must be
across it if the coil and condenser are to tune to 40 meters?
Fig. 94.—To what frequency will the
loud speaker and condenser tune?
116. Parallel resonance.—Many of the circuits used in radio
involve resonance in a branched or parallel circuit. Figure 95 shows
a typical parallel circuit composed of an inductance shunted by a
132 RESONANCE
condenser, the combination forming what is sometimes called an
“ anti-resonant ” circuit. The effects of varying the frequency of
the voltage across the circuit are widely different from the effects
in a series circuit. In the latter, the currents become very large
at resonance and the result-
ant series impedance of the
circuit becomes small. In the
parallel case the circuit offers a
large impedance to the genera-
tor and the current becomes
very small. In the series
case the same current flows
through the condenser and the
coil. The voltages across these
units differ. In the parallel
case the same voltage is across
each branch, but the currents
through them differ.
Experiment 3-7. Connect as in Fig. 95 the coil and condenser used in
Experiment 1-7. If sufficient meters are available read thea.-c.current in the
two branches as well as the current from the generator as the frequency of the
generator is changed. "Then fix the generator frequency and adjust the con-
denser capacity until maximum resonance occurs. Plot the currents against
frequency and against condenser capacity. ‘The generator in this experiment
may be a small oscillating tube. A 5-watt output is sufficient to produce cur-
rents in the branches of the circuit of 100 milliamperes which can be read with
a Weston Model 425 thermo-galvanometer.
In case laboratory apparatus is not available, the current may be calculated
after L, C, and H values have been chosen.
Ww
Fia. 95.—An anti-resonant circuit.
The same voltage exists across the branches and the circuit as
a whole. The current taken by each branch is the ratio between
the voltage and the reactance of that branch. Thus,
Se call
Tix ee
io = H/ Xe = HOw
— BY
ee 17 fete # (2 - cw) = p (=),
Lw Lw
Ue
PARALLEL RESONANCE 133
As the frequency isincreased, more and more current is taken by
the capacity branch, less and less by the inductance branch. In
the series case the voltages across coil and condenser are out of
phase; their algebraic sum at any frequency combined vectorially
with the JR drop is the voltage across the combination. In the
parallel case the currents are out of phase; at any frequency their
algebraic sum combined vectorially with the shunt resistance cur-
rent (if any) give the current taken from the generator. In the
- simple case where the resistance is neglected, the algebraic sum
of the currents gives the generator current. Since these two cur-
rents are out of phase (the capacity current has a negative sign),
adding them algebraically actually means subtracting Ic from Ix.
At resonance the currents taken by the two branches are equal
and if there is no resistance in the circuit the current taken from
the generator is zero, because it is the difference of the two branch
currents which is read in the generator circuit ammeter.
The impedance of the circuit as a whole, that is, the impedance
into which the generator must feed current, is the ratio between the
voltage and current as usual:
L= H/T.
Therefore, if no current flows, the circuit has infinite impedance.
Actually there is always some resistance in the circuit. This may
be in an additional shunt path, or it may exist in one or both of the
other branches. Actually, then, the generator current does not fall
to zero but passes through a minimum value. In most radio cir-
cuits by far the greater part of the resistance which is in the circuit
resides in the coil since the resistance of the average well con-
structed condenser used at radio frequencies is very small. In
such a circuit in which there is resistance in series with the induc-
tance and which is tuned to resonance by varying the value of the
capacity, C, the condition for resonance (minimum current from
the generator) is: ore jie
OT je eRe
The current taken by the circuit is not exactly in phase with
the generator voltage and so minimum-current resonance differs
slightly from zero-reactance resonance.
134 RESONANCE
ER
La R2 + wl?
and the impedance presented to the generator is
Wiel Te ib
[28 R :
117. Effective resistance.—If, as is usual, the resistance of the
coil is small compared to its reactance, the condition for resonance
is
and the minimum current from the generator becomes
ER
OP ree
w? L?
and since the impedance of the circuit is equal to the ratio between
the voltage across it and the current through it, it becomes
wD? L
R CR
Since at resonance the current into the circuit from the generator
is in phase with the voltage, E, across the circuit, the expression
above is not a true impedance but is more nearly a resistance.
It may be called the “ effective resistance ” of the circuit.
118. Resonant frequency.—In this manner we arrive at the
frequency to which a low resistance circuit becomes resonant:
RESONANT FREQUENCY 135
The condition for resonance then is that the inductive and capaci-
tive reactances are equal but opposite in sign—which is the same
condition that holds for series resonance. Here, however, there
are other conditions. The resistance must reside in the coil and
must be small compared to the reactance of the coil.
For example, in the case of the circuit in Fig. 95:
= 200 2 A
w = 27 X 356,000
R = 10 ohms (in the coil)
Xr = Lw = 200 X 10-* X 2 r X 356,000
= 447 ohms.
Here we may neglect the effect of resistance and use the simple
relation for resonant current and for impedance.
If E = 10
nn ER ER
B67 E27 (X22
TOS iG
~~ 4472
.5 X 10-3 amps. = 4 milliampere
Cale 4472
= = —— = 20,000 ohms:
and R i0
‘ : Lw
At frequencies other than resonance the impedance is re
a J (2 fo
provided there is no resistance in the circuit.
At lower frequencies than resonance most current goes through
the inductance because its reactance is low whereas that of the
condenser is high. As the reactance of the condenser decreases,
with increasing frequency, and that of the inductance increases,
and since the generator current is the actual difference between
these currents, the generator current decreases as resonance is
136 RESONANCE
approached. At low frequencies the circuit is said to be inductive
and ‘at high frequencies is capacitive.
119. Uses of series and parallel resonant circuits.— Whenever
it is desired to secure a large current and a low impedance circuit,
series resonance is utilized. When it is desired to built up a high
impedance or a high voltage circuit an anti-resonant circuit is used.
Let us consider the antenna-ground system in Fig. 96. The
antenna has in series with it an inductance
across which a voltage is to be developed
at a desired frequency. In series with this
inductance are a capacity for tuning purposes
and an anti-resonant circuit. Voltages of
various frequencies, among them the desired
frequency, are impressed on the antenna by
distant transmitting stations. The maxi-
mum voltage is desired across the induc-
tance, L, at the desired frequency and the
minimum at other frequencies. There is a
specially strong signal which is setting up
a voltage across the antenna. The anti-
resonant circuit is tuned to this frequency.
The condenser C is adjusted until the
antenna system as a whole is resonant to
= the desired signal. A large current flows
Fic. 96.—The anti-reso- through the series system, building up a
nant circuit in series Jarge voltage across L. Voltages of other
ears “oetn a Telectey rr uencies cause small currents to flow i
undesired signals by q oe MEDC DES 20. RONG
making the series im- the antenna system and consequently small
pedance to them very Voltages at these frequencies are built up
high. across the coupling coil, LZ. The anti-
resonant circuit’s being tuned to the un-
wanted signal makes the antenna system as a whole have a very
high impedance at this frequency and so very small currents will
flow through it, building up small voltages of this frequency
across the coupling coil.
Such an anti-resonant circuit is often called a rejector circuit
because it rejects signals of the frequency to which it is tuned.
SERIES AND PARALLEL RESONANT CIRCUITS 137 -
The series resonant circuit is called an acceptor because it accepts
signals of the resonant frequency. The rejector used in this cir-
cuit is commonly known as a wave trap because it traps out un-
wanted signals.
Let us suppose signals are fed into the input of an amplifier
which has an internal resistance, R, which is in series with an out-
put circuit, as shown in Fig. 97. The voltage across this output,
Z, is to be made as high as possible. The amplifier has available
a certain voltage, H, which must be divided between the internal
resistance of the amplifier and the output load. The proportion of
the voltage that appears across this load increases as its ohmic
Amplifier
30010 °H
R=Internal Res L
: wo
of Amplifier WHEE Cr
f=1000 KC
cC=1000x10 " F
Fic. 97.—A high impedance is desired for the amplifier to work into. A tuned
circuit does the trick. Numerically it is equal to Z.
impedance increases with respect to the amplifier’s resistance.
Thus, if the output impedance is equal to the internal resistance
of the amplifier, one-half of the total voltage available will appear
across it. If it is higher than this value, a greater proportion of
voltage will be usefully applied across the load and less used up in
the resistance.
In this case the anti-resonant circuit is used. At resonance its
L Dw?
impedance becomes equal to CR or >
(300 X 10-%)2 X (6.28 X 1000 X 1000)?
20
= 180,000 ohms.
If the amplifier’s internal resistance is equal to 20,000 ohms,
the voltage across the tuned circuit is $$$ or 7% of the total avail-
able voltage.
y
\ |
yf \
Aj ,
Cv
1385 oe hae RESONANCE
Problem 16-7. A screen-grid tube gives the greatest voltage amplification
when worked into a very high impedance. A condenser of 1500 mmfd. is
available. Calculating the size of the inductance required to tune to 1000
meters and assuming it has a resistance of 30 ohms, what is the impedance
26y2
(= that can be presented to the tube by shunting the coil and condenser?
*:
Problem 17-7. A wave trap is to be put into an antenna and tuned to a
station whose frequency is 750 ke. What will be a convenient size of con-
denser and coil to use? They are to be shunted across each other and the
combination put in series with the antenna. If the coil has a resistance of
10 ohms and the condenser a resistance of 1.0 ohm at this frequency, what
impedance will the trap offer to the offending signal?
Problem 18-7. In Problem 17-7, neglecting phase differences between
the trap and the rest of the antenna, if the total impedance of the antenna to
the offending signal is double that of the trap alone so that one-half of the
total antenna voltage is across the trap, what current will flow through the
condenser if the total 750-kc. voltage across the system is 10 microvolts?
120. Sharpness of resonance.—The effect of resistance is to
reduce the maximum current flowing in a series resonant circuit,
and to make less pronounced the minimum of current flowing into
a parallel resonant circuit from an external source.
Since the maximum current is desired in a series circuit, and the
maximum impedance in a parallel case, the inclusion of resistance
in either is deleterious.
Let us consider the antenna illustrated in Fig. 96. Suppose its
inductance, L, is 200 microhenrys and C at resonance (356 kc.) is
1000 mmfd. For the moment we shall neglect the presence of
the wave trap. Assume a voltage of 10 volts. What is the
effect on the resonance curve of this antenna system if it has a
resistance of 10 ohms or of 40 ohms? The current at resonance in
the 10-ohm case is 1 ampere whereas at 370 ke. the current is
.274 ampere, a ratio of 3.65. In this 40-ohm case, the resonant
current would be only 0.25 ampere—one-fourth of its value with
the lower resistance—and the current at 370 ke., ie., 14 ke. off
resonance, would be .188 ampere. This is a current ratio between
the resonant and the off-resonant current of only 1.33.
In other words, if the antenna had impressed on it from equally
distant and equally powerful radio stations two voltages, one of
356kce.—the desired frequency,—and one of 370ke.—the unwanted
SHARPNESS OF RESONANCE 139
frequency,— 3.65 times as much current flows at the desired fre-
quency as the unwanted. In the 40-ohm antenna, however, not
only is the desired current cut to one-quarter of its other value
but the ratio of wanted to unwanted current has been decreased
to 1.33. The low-resistance antenna is said to be more “ selective ”
and its “ selectivity ’’ is decreased when resistance is added to it.
121. Selectivity.—The selectivity of a circuit is a measure of
its ability to distinguish between wanted and unwanted signals.
The steepness of the resonance curve is a direct measure of this
selectivity.
Let us consider the parallel or anti-resonant circuit. At its
resonant frequency it keeps currents of undesired frequency from
flowing through the antenna because of its high impedance at
those currents. This impedance, L?w?/r, increases as the resist-
ance of the circuit decreases, so it behooves the designer to use
low-resistance coils and condensers when building a trap or rejector
circuit.
Since a circuit may be tuned to resonance by varying any one
of three variable factors, the inductance, capacity, or frequency,
we may express the sharpness of resonance in any one of three
ways. It may be the fractional change in current for a given frac-
tional change in either L or C. Naturally the sharper the resonance
curve and the greater its height, the greater will be the current
change for a small number of degrees of change in the tuning con-
denser. The circuit will tune ‘“‘ sharply’; it is called a sharp
circuit. In practice the condenser is used as the tuning variable.
If, then, the current at resonance J, and the tuning capacity C, are
noted and then changed to give some other value of current, the
sharpness of resonance may be found by substituting values in the
following expression,
By some mathematical juggling of this cumbersome expression
(see Bulletin 74, Bureau of Standards, page 36) a much simpler
140 RESONANCE
expression may be obtained. This has two forms,
ik La
Re i Ree
Sharpness of resonance =
where R = the resistance of the circuit;
C, = the capacity at resonance;
L = the inductance of the circuit.
l|
In other words the sharpness of resonance is the ratio between
the capacitive or inductive reactance to the resistance, and thus
the resonance curve rises the steeper the less resistance there is in
the circuit.
Another expression for the sharpness of resonance is obtained
by varying the frequency and noting how the current changes.
Thus an expression is worked out which shows the width of the
resonance curve where the current is equal to .707 X I,, where
I, is the resonant current.
Suppose, as in Fig. 98, we plot a resonance curve of current
against capacity. Suppose the capacity is adjusted until the total
reactance in the circuit (Xz — Xc) is equal to the resistance in the
circuit. That is,
(Xz FY Xc) = Rk,
E E
when I = —— becomes equa! to
V R2 + X? : NV 2|
and ae Oe
Lw 2G; Ww
then Bie CRG: yen.
in which C, = the capacity at resonance;
C; and C2 = the two values of capacity which makes I = .707 I,.
122. Width of resonance curve.—If, however, the frequency of
the impressed voltage is so adjusted that two currents are reached,
WIDTH OF RESONANCE CURVE 141
above and below the resonant frequency f;, which are equal to
pacdiel,,
Lo _ _ fr
RK fe-—fi
whence the width of the frequency band
=H
(Pee
900 925 950 975 1000 1025 1050 1075 1100
C Mmfd.
Fra. 98.—If J,, and I are equal to 0.707 times the resonant or maximum cur-
rent, the resistance of the circuit may be calculated.
Example 4-7. What will be the width in cycles of the resonance curve at a
point where J = .707 J, when L = 200uh, R = 10 ohms, f = 356,000 cycles?
_ 10 X 356,000 _ R X fr
hi-fi 447 Lew
8000 cycles
142 RESONANCE
dif Fe
and 1 Re. COCs
LIGRE
Le
_ 2000 X 10 |
«447
= 44.7 mmfd.
= change in capacity required to change the
current from J = .707 I, below resonance, to J = .707 J, above resonance,
or from J, to J, in Fig. 98.
/ Problem 19-7. In Fig. 98, suppose the resistance is 20 ohms instead of |
10. Calculate the width of the band at the point where the current is 0.7 of
its maximum value, and the change in capacity required to produce this
change in current.
Problem 20-7. A certain esil-condenser combination has a resistance of
16 ohms at 400 meters. The inductance is 170 microhenrys. What is the
width of band passed at the point where the current is equal to 0.7 of its
maximum value? What is the discrimination between the resonance current
and another current 10 ke. off resonance? What is the “ sharpness of reso-
nance ” of this circuit?
Problem 21-7. A circuit is to pass only 0.707 of its maximum current at
a point 2.0 ke. off resonance, which occurs at 500 ke. The condenser to be
used has a capacity of 0.0006 mfd. Calculate the maximum resistance the
circuit can have.
Problem 22-7. Suppose that increasing the size of an inductance by a
factor of 2.0 increases the resistance in a circuit by a factor of 1.5. The circuit
is to tune to the same wavelength. What has happened to the sharpness of
resonance, or, what amounts to the same thing, to the selectivity of the
circuit?
~ Problem 23-7. If the expression Lw/r of a coil remains constant over a
fairly wide band of frequencies, does the selectivity of a tuned circuit differ at
different frequencies? Does the width of band passed differ at 1500 ke. from
what it is at 500 ke.?
The selectivity of a radio receiver can be illustrated by Fig.
188 which shows the relative gain of a stage of a radio fre-
quency amplifier when the signal is so-and-so-many kilocycles
away from the frequency to which the receiver is tuned. Note
the sharpness of the curve at 550 ke. and the poor selectivity
at high frequencies.
EFFECT OF INDUCTANCE 143
123. Effect of inductance and capacity on sharpness of reso-
nance.—Since the sharpness of resonance expression, Lw/R and
1/wCR both show that the inclusion of resistance tends to cut
down the selectivity of the circuit in which the resistance exists
it behooves the experimenter and engineer to keep the nn
of his circuits at a minimum—when selectivity is his goal. What
effect has changing the ratio of inductance to capacity, the product
of L X C remaining constant?
‘ear
9.0 ,
L=100 nH
8.0 am f.. 1 - / \ C=2000 a
\
7.0 , | 1
6.0
10.
—
4.0
| L=4001H |
20 C=500 Mmfd. | |
| ¢=1000 Mmfd:
x) | L=200”H | L E=100 l
Z|
330) 335) 340) 3455350" 355 360! 365 9370) 375
Fra. 99.—Effect on sharpness of resonance of varying ratio of L to C.
1.0 7
Let us consider the ratio of inductive reactance or resistance,
Lw/R. Tf we can increase L without increasing R we shall increase
the sharpness of resonance. Now considering the ratio of capaci-
tive reactance to resistance, 1/CwR, increasing C has the same
effect as increasing the resistance—the sharpness of resonance is
decreased, the selectivity of the circuit goes down.
In a series circuit, then, the selectivity increases as the ratio
L/C increases. Some theoretical curves showing this effect are
plotted in Fig. 99 showing that for selective circuits a large induc-
tance and small condenser should be used. (R = 10 ohms.)
144 RESONANCE
In a shunt circuit the opposite is true, the selectivity increases
as the ratio of L/C decreases. In other words, for a selective
parallel tuned circuit, a large capacity and small inductance should
be used. Some curves showing this effect are shown in Fig. 100.
In both of these cases, the product of L X C must be a
constant in order that the same resonant frequency be main-
tained.
124. The resistance of coils.—The difficulty with the adjust-
ment of one’s circuits to the desired selectivity by using large coils
ace
330 340 345 350 355 360 370 380
Frequency in KC
re 25
Fic. 100.—How varying C affects sharpness of tuning of an anti-resonant
circuit.
and small condensers or vice versa according to the above section
lies in the fact that the resistance of a coil increases as the induc-
tance is increased and the curves in Figs. 99 and 100 were plotted
on the assumption that the resistance of the circuits in question
remained constant. Such is not the case, and so the increase in
sharpness of resonance by using a large ratio of inductance to
capacity (in a series circuit) is not so pronounced as theory would
indicate.
The question naturally arises, is the resistance of a coil different
at different frequencies, and if so, why?,
HIGH-FREQUENCY RESISTANCE 145
The answer is that the resistance of a coil of wire to high-fre-
quency currents may be several times its resistance to d.-c. currents.
The 200-microhenry coil used in examples several times in this
chapter which at 356 ke. has a resistance of 10 ohms is a good coil,
and yet its resistance to direct current is probably an ohm or less.
125. High-frequency resistance.—In all alternating current
problems the resistance that is considered is the resistance at the
frequency under consideration. Thus at broadcast frequencies
550 to 1500 ke., a coil will have a certain a.-c. resistance; at 60
cycles its resistance will be different, and to d.-c. currents its
resistance may be still another figure.
A wire stretched out straight will have one resistance to d.-c.
and another to a high-frequency current; therefore the fact that
the wire is coiled up in an inductance is not the cause of the addi-
tional resistance. The difference arises from the fact that the
current in a conductor at high frequencies is not evenly distributed
throughout the cross-section of that conductor. Because of the
rapid change of direction of flow and because the current within
the cross-section of a conductor changes rapidly, small e.m.f.’s are
generated in that cross-section, and therefore all along the wire.
These voltages are in such a direction, according to Lenz’s law
(Section 55), that of the total current flowing more is along the
surface of the wire and less along the inner parts of the wire. The
result is a decrease in effective area of conductor and a consequent
rise in resistance.
A table is given in Circular 74 (Bureau of Standards) showing
the effect of diameter of wire, frequency, resistance, etc., upon this
phenomenon known as “‘ skin effect.”” For our purposes it is suffi-
cient to know that the resistance of a coil to high-frequency cur-
rent is always greater than its resistance to a direct current.
The resistance of a coil over the range of frequencies at which
it is used changes somewhat, increasing with increase in frequency.
The manner in which the expression Lw/R of a coil, sometimes
called its “Q,” varies over the frequency range is plotted in Fig.
101. Knowing this factor for the coil in a series or shunt circuit
we can calculate the width of the frequency band at a point where
the current is 0.707 of its resonant value, we can plot a resonance
146 RESONANCE
curve, and can calculate the equivalent impedance of the circuit
at resonance to a generator which must feed current into it.
Let us, however, measure the resistance of the coil at higher and
higher frequencies. What happens? Figure 102 shows that at
higher frequencies the coil resistance becomes very high and finally
the curve rises perpendicularly, indicating that at some nearby
point the resistance is infinite. What is happening?
126. Distributed capacity of coils—Whenever two objects
which conduct current are insulated from each other, they form
a condenser. Electricity may be stored in it. Its capacity depends
200 4 +—
=O
5 150 | ae |
> a
100
500 600 700 800 900 1000 1200 1400
Frequency in K.C,
Fic. 101.—How the “Q” (Lw/r) of a coil varies with frequency.
upon the proximity of the objects, the insulation between them, and
their shape. In a coil of wire each turn is at a different potential
from its neighbor, and is separated from it by the insulation of the
wire. Thus every coil is not a pure inductance but may be thought
of as a coil shunted by a capacity made up of the resultant capacity
of a number of smaller capacities. At some frequency the coil
shunted by its capacity becomes anti-resonant, and the circuit
then becomes as shown in Fig. 103 where the tuning condenser is
no longer in series with a coil but with a parallel tuned circuit
which at the resonant frequency has a very high impedance. The
DISTRIBUTED CAPACITY OF COILS 147 ©
series impedance of the circuit, then, increases as we approach the
resonant frequency (sometimes called the natural wavelength or
12
11
10
Res. Ohms
L= 1252 uh
100 200 300 400 500 600 700 800 900 1000
Wave Length Meters
Fig. 102.—High-frequency resistance of a coil.
frequency) of the coil, and for this reason its effective resistance
becomes great.
148
RESONANCE
This capacity of the inductance is known as a distributed
Fig. 103.—When the dotted capacity
across the coil tunes it to the fre-
quency of the generator, the series
impedance of the circuit becomes
very high.
capacity since it is not concen-
trated in any one place or form
but is more or less evenly dis-
tributed along the whole length
of the inductance. Its effect is
to lower the effective inductance
of the coil and to increase its
resistance somewhat. The ca-
pacity and inductance of a coil
do not change much with fre-
quency, but the apparent induc-
tance of this anti-resonant circuit does change with PEMA
It is equal to
L
sl ecoce
in which C, = the capacity of the coil;
De —
ll
its apparent inductance;
its true or low frequency inductance;
6.28 X f.
CHAPTER VIII
PROPERTIES OF COILS AND CONDENSERS
Corts and condensers form the nucleus of every radio circuit.
Other apparatus is needed, of course, but for each of the other
units needed there are several substitutes. There are no substitutes
for coils and condensers. To understand what their rdle is in the
reception of radio messages, either in code or voice or musical form,
we must look at a simple receiving system.
127. Tuning a receiver.—A simple receiving circuit consists of
an antenna-ground system connected to a coil and a “ detector ”
such as a crystal of carborundum or galena or silicon or other
sensitive mineral which has the property of separating the audio
tones from a radio wave. A pair of head phones may be put in
series with the detector so that the audio tones
which are filtered out of the radio wave by the
detector may be made audible. A small condenser
across the phones will pass the radio frequencies
but not the audio frequencies which must go through
the phones.
One way to get louder signals is to tune the
antenna-ground system to the frequency of the .
desired wave. This is done by varying C in Fig. ae
104. When the circuit is series resonant, a large ae
current flows through the inductance. The volt- ,. 104.—A
age across it (Xz X I) will be large and the re- gimple radio
sponse from the crystal will be greater. receiver.
The voltage across the inductance can be ampli-
fied and then impressed across the detector. This amplification
may take place in several stages so that very weak signals may
149
rc
150 PROPERTIES OF COILS AND CONDENSERS
finally be heard with the strength of nearby strong signals
which are detected directly from the antenna inductance. If
desired, the signals may be amplified again after detection by
means of audio-frequency amplifiers.
As we have already seen (Section 121), there is another advan-
tage of tuning the antenna, the advantage of selectivity. Signals
of low frequency find considerable impedance in the condenser of
such a series-tuned antenna, signals of high frequency find impe-
dance in the coil; signals of the desired or resonant frequency find
a minimum of impedance, and so the filtering action of the tuned
system is advantageous. If, in addition to
the series tuned circuit, we used an anti-
resonant or parallel-tuned circuit as in Fig.
105, we impose more hardships upon un-
wanted signals. In this case when maximum
Ci current flows through L; maximum current
C, is induced in Ly. If, then, C2 is tuned so
L, eee that L2C2 form an anti-resonant circuit, the
impedance to the resonant frequency will be
als very high and any current through it will
Poe var pee a large voltage across it so that the
Gnd the antenna yee eetor gets a high voltage at the desired
tem as a whole is series frequency and a low one at all other fre-
resonant increases volt- Quencies—and the selectivity of the system
age across Ly. as a whole is improved.
If, in addition, each radio-frequency am-
plifier stage is tuned to the desired signal, the selectivity of the
entire receiver may become very great. In the present congestion
of broadcast stations, the necessity for selectivity of a high degree
is evident; as we shall see later it is a disadvantage.
128. The wavemeter.—An instrument for measuring the wave-
length or frequency of signals is called a wavemeter when calibrated —
in meters or a frequency meter when calibrated in kilocycles or
cycles. It consists of a coil and a condenser and some means of
indicating when this simple circuit is tuned to resonance with a
radio wave. The indicator may be a current meter, a lamp which
lights up at maximum current through it, or a crystal detector and
HETERODYNE WAVEMETER 151
a d.-c. milliammeter. It may be connected directly into the cir-
cuit, or, preferably, coupled loosely to it.
The circuit of a simple and effective wavemeter is shown in
Fig. 106. The indicating device is a crystal detector and a meter
which indicates the rectified d.-c. current. If it is coupled loosely
to the tuned circuit the resistance of this indicator will not broaden
the response curve of the wavemeter. The inductance is usually
fixed and the capacity varied to obtain resonance, but to cover a
wide band of frequencies it is frequently necessary
to have several coils which fit into the wavemeter
by means of plugs and jacks. If the coils are
arranged so that the larger coils have exactly four
times the inductance of the next smaller the wave-
length range will be doubled, and the frequency
range halved.
A series of coils in which the same winding
7
Fie. 106. — A
space is used but in which the number of turns in
this space is doubled for each next larger coil will
approximate very closely these conditions.
Sometimes the wavemeter is equipped witha
buzzer so that it will send out a modulated wave.
wave meter in
which the resist-
ance of the indi-
cator is removed
from the tuned
circuit and is
A receiver can be tuned to a desired frequency
by starting the buzzer, tuning the wavemeter to
the desired wavelength, or frequency, and ad-
justing the receiver until the buzzer tones are heard at maximum
loudness.
129. Heterodyne wavemeter.—The most useful type of wave-
meter is the heterodyne wavemeter which uses an oscillating
vacuum tube and meter, usually in the grid circuit. The circuit
diagram for such a meter is shown in Fig. 107. Tube 1 generates
radio-frequency currents, which are modulated when desired by
the low- or audio-frequency generator tube 2. Such a meter gives
very sharp indications of resonance, and because it is a small
modulated source of radio-frequency energy it can be used to tune
receivers to any desired frequency. It is a much more accurate
instrument than the buzzer wavemeter. The data in Table I are
those of an oscillator-wavemeter used in Radio Broadcast Labora-
coupled loosely
to it.
152 PROPERTIES OF COILS AND CONDENSERS
tory. The coils are standard General Radio Company inductances
(30 to 1000 meters).
Plug we
+12) ——we AL
0.0005
—A
f
i y
ee
Radio Frequency
R.F. Choke
Q0QQ00000
0.0012
0.
a
Gy
Audio Frequency
Fic. 107.—Circuit diagram of a heterodyne-wave meter or modulated
oscillator.
TABLE I
Coil » f Ke. per
ce : dial degree
15 45-120 2500-6660 31.6
30 80-210 1430-3750 OSS
60 165-400 750-1820 LOR
90 265-620 485-1130 Gu5
Coil Turns Size Wire | Diameter Tener) oO If,
Winding
277-A 15 2h 24 13 014 mh
277-B 30 21 2H 13 055 mh
277-C 60 21 21 13 217 mh
277-E 90 27 24 13 495 mh
130. Calibrating a wavemeter.—A wavemeter, or frequency
meter, to be most useful must be properly calibrated. This may
be done in several ways. If the meter is a heterodyne meter all one
needs is a source of known frequency and a receiver. The process
CALIBRATING A WAVEMETER 153
issimple. Tune the receiver to a station whose frequency is known.
Then turn on the oscillating tube wavemeter, and when a whistle
is heard from the receiver, the known station, the receiver, and the
wavemeter are all tuned to the same frequency. Then tune the
receiver to another frequency and repeat the performance. Then
a curve can be plotted showing the calibration of the wave meter.
The following description of how to calibrate a wave meter
over a wide range of frequencies by means of but a single accurately
known frequency is an interesting experiment. It follows from the
fact that an oscillating vacuum tube generates not only the fre-
quency governed by the LC product of its circuit but also multiples
(harmonics) of this frequency.
Experiment 1-8. . To calibrate a wavemeter by harmonics.—The necessary
apparatus consists of:
(1) An oscillating wavemeter connected as in Fig. 107.
(2) An oscillating detector tube preferably followed by a stage of audio
amplification.
Tune the oscillating detector to the frequency of some known station by
listening in the head phones and bringing an antenna wire near the detector
inductance. The condenser of the detector should be equipped with a vernier
or worm gear so that very accurate settings are possible. Tune as nearly as
possible to ‘‘ zero ”’ beat with the known station. As the tuning dial is adjusted
near resonance with the known station, now acting as our frequency standard,
a note will be heard in the phones which represents the difference in frequency
between the known station and that of the detector tube. When this difference
tone (or beat note) disappears, the two oscillations are at the same frequency.
Since frequencies lower than about 100 cycles cannot be heard in the phones,
it will not be possible to tune closer than this to the desired frequency. By
estimating the two points at which the audible beat disappears and finally set-
ting the oscillating receiver detector at the mid-point between these two dial
settings, a sufficiently accurate setting will be made.
We have now equipped outselves with a local generator whose
frequency is accurately known. For example, suppose it is 610 ke.
and that we are set to within 100 cycles of this frequency. We are
within 100 parts in 610,000 of being exactly correct or one part in
6100, which is sufficiently accurate. It is much more accurate
than we can read the dial on the wavemeter we are to calibrate.
Now move away the antenna coupling and see if the beat note
changes. If it does, again adjust for true zero beat. Then start
154 PROPERTIES OF COILS AND CONDENSERS
up the oscillating-tube wavemeter and, after giving it a few
minutes to warm up, tune its dial slowly until a whistle or beat note
is heard in the head phones which are still plugged into the detector-
amplifier. This means that the wavemeter is being tuned to the
frequency of the oscillating tube.
If we use the broadcast band coil of the wavemeter we ought to
get a very loud beat note when the two circuits are in exact reso-
nance and another loud note when the dial is tuned to the half
wavelength, in this case 1220 kc. In between these points may be
several other weaker beat notes.
TABLE II
Dial aR Units f—approxi-
Degrees ea: Difference mate ete
L032 Sgn seca iesreaere | crus terete cuter 1220 1220
34.0 23.8 2 1020 1016
47.0 13.0 1 920 915
60.0 13.0 1 820 813
85.0* 25.0 2 610 610
Now turn the dial slowly and put down on paper each time a
beat note is heard. For example, the table of such points may
look like Table II, in which the loudest beat notes are marked
with an asterisk. Then use another wavemeter coil and repeat,
always marking down the loud notes.
Now prepare data like those in the next table, in which the
numbers along the top are obtained by multiplying the detector
frequency by whole numbers from 1 to 10, and the vertical num-
bers are obtained by dividing this frequency by whole numbers.
Thus our fundamental frequency is 610 ke. Twice this is 1220 ke.,
one half is 305, ete.
Then make a list from this table of the frequencies that may be
looked for from our calibration, namely: 610, 763, 813, 915 ke., ete.
What actually happens as we tune the wavemeter dial and hear
beat notes? The oscillating detector and the wavemeter tubes
CALIBRATING A WAVEMETER 155
are generating additional or harmonic frequencies as well as the
fundamental to which they are set. These additional frequencies
are much weaker than the fundamental. When we tune the
wavemeter to 1220 ke. it beats with the second harmonic of the
detector and gives an audible note. But how are we to recognize
the 1220 point? How do we know it is not the third or the fourth
harmonic instead of the second?
Consider the data in Table II. We got loud notes at 10.2°
and 85°. We guess that these are the second harmonic and funda-
mental. We subtract the dial settings as in column 2. Then
assuming that 13° is a unit, we note that there are two units
between the 10.2° and the 34° beat notes. We see then that there
are six units between 1220 and 610 kc. We guess again and say
that each beat note represents about one-sixth of the difference
between 1220 and 610 ke., or about 100 ke. per unit. Looking in
our list of expected frequencies we can pick out these frequencies
exactly.
TABLE III
1 2 3 4 5 6
1 610 1220 1830 2440 3050 3660
2 305 610 915 1220 1525 1830
3 202.5 406 610 813 1016 1220
4 152.5 305 457 610 763
5
6
We might guess at these frequencies from the original assump-
tion that the two loud notes were from the 1220 and the 610 ke.
frequencies and noting that between them—a difference of 610
ke.—were 85—10.2 dial degrees or about 8 ke. per degree.
When the smaller coils are to be used, care must be taken to
see that no harmonics are missed. Fortunately, if the coils have
the dimensions given in Table I, the harmonics will fall at almost
the same points on the dial. Thus on the largest coil 610 kc. is
found at 85°. On the next smaller coil the 1220 ke. frequency will
156 PROPERTIES OF COILS AND CONDENSERS
be found within a degree or two of 85°. And so on until the entire
set of coils is calibrated.
131. Standard Frequencies.—In this country standard fre-
quency signals are sent out from the Bureau of Standards at stated
intervals and can be heard at distances up to 1000 miles from
Washington, D. C. In addition there are many long-wave and
intermediate-wave stations whose frequencies are kept within very
close limits and which are ‘‘ on the air ”’ 24 hours of the day. The
broadcasting stations themselves form good standards of frequency
—especially the better known stations—covering the band from
550 to 1500 ke., and above this are many short-wave stations
whose signals may be heard the world over.
132. Calibrating by “‘ clicks.””—A method of calibration that
is often used is the click method. When a tuned circuit is brought
near the inductance of the heterodyne wavemeter, a sharp dip of
the grid current needle will be noted as the two circuits are reso-
nated to each other. If the same tuned circuit is brought near the
inductance of an oscillating detector tube, a sharp click will be
heard when the circuits are tuned to the same frequency provided
one listens in the plate circuit of this tube or behind a stage of audio
amplification. This click is produced by a sharp change in grid
current and a corresponding change in plate current.
Experiment 2-8. Calibration by clicks.—This method requires an oscil-
lating detector, a standard meter and the unknown meter to be calibrated.
Couple the standard meter to the inductance of the detector, and turn
the dial until a sharp click is heard in the phones indicating that the circuits
are tuned alike. If the two inductances are closely coupled two clicks will
be heard, one when the tube stops oscillating and one when its starts again.
These two points may be several degrees apart. Loosen the coupling, and
note that the two clicks approach each other. Keep on loosening it until a
degree of coupling is reached when only a single resonance click is noticed.
Note the dial setting of the standard meter. Now remove it from the tuned
circuit and bring near the latter the wavemeter to be calibrated. Turn its
condenser dial until a click is heard as before. Now the meter has the same
frequency, or wavelength, as the standard. Other points for a calibration
curve may be noted in the same manner.
This method really constitutes setting a generator or miniature transmitter
(the oscillating detector) to a given frequency by means of the standard meter
and then tuning the uncalibrated meter to resonance with this generator.
THE PROPERTIES OF COILS AND CONDENSERS 157
133. The properties of coils and condensers.—We may ‘investi-
gate the properties of coils and condensers by performing the
various parts of the following experiment.
Experiment 3-8. Wind up on a form about 3 inches in diameter, a coil
of about 60 turns of rather large wire, preferably with silk or enameled insula-
tion so the distributed capacity of the inductance will be rather large. Connect
it across a condenser whose maximum capacity is about 500 mmfd. Starting
R 2
130,000
L= 3:54C
= 130,000 — 40,000
~~ 3.54(196—38)
= 232uh
Distributed
Cap.=32 mmfd
40 20 0 20 40 60 80 100 120 140 160 180 200
C° mmfd
Fic. 108.—A method of determining the distributed capacity of a coil.
at the maximum capacity of the condenser, measure the resonant frequency of
the coil-condenser combination by “ clicking ” it into an oscillating receiver,
or by coupling it to an oscillator. ‘Then decrease the capacity and repeat until
several readings are taken, say at 500, 400, 300, etc., mmfd. Plot the result
against C as shown in Fig. 108; that is (wavelength)? against capacity.
A straight line results because the formula
(Wavelength)? = 3.54 L X C,
where DL is in » h and C in mmfd.,
158 PROPERTIES OF COILS AND CONDENSERS
is the equation of a straight line and states that the wavelength squared is
proportional to the capacity in the circuit. The slope of the line divided by
3.54 is the inductance of the coil, that is,
aw Ned
= sare 1G
It will be noticed that the straight line crosses the wavelength squared
axis at some distance above the zero point. This gives us the natural wave-
length squared of the coil itself and therefore the resonant wavelength to
which the coil with no additional capacity will tune. The point where the line
crosses the capacity axis gives us the distributed capacity of the coil. This
value multiplied by the inductance as obtained above gives the LC product
which when fitted into the proper formula gives the natural wavelength of the
coil.
Thus, in one experiment we can determine not only the frequency or wave-
length to which a coil-condenser combination will tune, but we can determine
the coil’s inductance, its distributed capacity, and its natural wavelength.
As a check on these data: (a) calculate the inductance from the formulas
given in Fig. 51. (b) If a heterodyne wavemeter is available and calibrated
to short wavelengths, detach the condenser from the coil and click the latter
into it, and thereby determine the natural wavelength of the coil.
L
134. Measurement of coil resistance.—The effect of resistance
upon the sharpness of resonance and the selectivity of the circuit
has been mentioned (Section 112). The resonance curve gives us
one method of measuring the resistance in a given circuit, provided
we know the inductance of the coil—which can be calculated from
the formula in Fig. 51.
Experiment 4-8. To determine the resistance of a coil—Couple a series
circuit composed of a coil, condenser, and indicating meter to a generator of
about 5-watts output. Adjust the frequency of the generator through reso-
nance with the series circuit. If the generator has a constant output over
this frequency range the accuracy with which the coil resistance is determined
will be greater. Pick out the two frequencies above and below resonance
where the current in the circuit is .707 of its value at resonance and calculate
the width of frequency band at this point and the resistance of the circuit,
from the equation
_ Le (is =f)
Sr
Subtract from this value the resistance of the current meter.
For example a model 425 Weston thermo-galvanometer will read
currents of 115 milliamperes and has a radio-frequency resistance
R
MEASUREMENT OF COIL RESISTANCE 159
of 4.5 ohms. The value of resistance remaining is the resistance of
coil, leads, and condenser. Most of this resistance resides in the
coil.
Experiment 5-8. To determine resistance of a circuit.——Another method
of determining the resistance of a coil is as follows. It necessitates the use of
a decade resistance box or series of accurately known resistances of negligible
inductance and capacity and a variable condenser.
Small lengths of high-resistance wire (manganin) are to be preferred for
frequencies higher than 1000 ke. Their d.-c. and high-frequency resistance
is practically the same.
Connect the apparatus in series and couple to an oscillator.
With the resistance box short-circuited (R = O), tune the circuit to reso-
nance. Then add enough resistance to the circuit to halve the current,
retuning to resonance, if necessary. Then since we have halved the current,
Ohm’s law tells us that we have doubled the resistance. In other words the
added resistance is equal to the resistance already existing in the circuit.
Again subtract the resistance of the current-indicating meter. What remains
is the resistance of coils, condensers and leads.
Repeat at several different frequencies and calculate the
resonance,” Lw/R, and plot against frequency and wavelength.
“sharpness of
If only one or two resistance units are available, say 5 or 10
ohms and not a continuously variable standard of resistance like
a decade box, the resistance of the circuit above may be determined
by noting the current at resonance, and the current when some
resistance has been added, retuning to resonance after adding the
resistance if necessary. Then the current, according to Ohm’s
law, is
E
i= R
ee E
OS 1p Ee ey
where J, = current at resonance and no added resistance;
Ie = current at resonance and 2 added;
R, = resistance of circuit;
Re = added resistance;
Role
whence Ry
I; — le
160 PROPERTIES OF COILS AND CONDENSERS
If a current-indicating meter is used whose deflections are
proportional to the current squared, an example is a thermo-
galvanometer or a hot-wire meter, it is only necessary in this
experiment to add sufficient resistance to quarter the deflection of
the instrument. This is equivalent to halving the current, and
the added resistance is equal to the resistance already in the circuit.
The lower the resistance of the current-indicating device, the
greater will be the accuracy with which such measurements may be
carried out. For example, if the indicator has a resistance of 4.5
ohms and the circuit a resistance of 5 ohms, great accuracy cannot
be attained, but if the circuit resistance is double or treble that of
the indicator, much greater accuracy results. In any case the
meter resistance must be subtracted from the measured resistance
to get the resistance due the circuit alone.
135. Condenser capacity.—We will now investigate by means
of an experiment the capacity of a condenser.
Experiment 6-8. To determine the capacity of a condenser.—Connect a
variable condenser whose calibration is known across a coil and click into an
oscillating receiver or into a heterodyne wavemeter; attach the unknown
condenser across the variable condenser and retune the latter to resonance
with the wavemeter. The difference in readings of the calibrated condenser
is the capacity of the unknown condenser. For example, suppose resonance
is obtained by the variable condenser alone when set at 400 mmfd. Connect-
ing the second condenser across the variable forces us to reduce the capacity
of the latter to 8340 mmfd. The difference 400 — 340 = 60 mmifd. is the
capacity of the unknown. Such a method enables the experimenter to dis-
regard the capacity of the coil itself or of the leads since these are connected
across the variable at all times and do not change when the unknown is
attached to the circuit.
136. Antenna wavelength.—We will proceed to determine the
wavelength of an antenna by means of the following experiment.
Experiment 7-8. To measure the natural wavelength of an antenna.—
Connect in series with the antenna an inductance which can be adjusted in
even steps, say a coil of 20 turns with taps at each turn. Measure the fre-
quency to which the antenna tunes with the entire coil in the circuit by coupling
the coil to a heterodyne wavemeter. Then reduce the inductance by one turn,
and repeat. Repeat until accurate readings are no longer possible. Plot
wavelength, or frequency, against added turns of wire. Where the line crosses
ANTENNA INDUCTANCE 161
the wavelength or frequency axis is the natural wavelength or frequency of
the antenna.
137. Antenna capacity.—The capacity of an antenna may also
be determined by experiment.
Experiment 8-8. To measure the capacity of an antenna.—Measure the
wavelength of an antenna attached to an inductance, as in Fig. 109. Then
replace the antenna-ground connections
by a variable condenser (Fig. 109b) and VY
tune the condenser until resonance with !
the wavemeter is indicated. The capacity :
of the condenser at this point is the '
capacity of the antenna.
138. Antenna inductance.—Ex- 3
periment is resorted to to determine
the inductance of an antenna. ph
Experiment 9-8. To determine the
inductance of an antenna.—Connect a
known inductance, Li, in series with the =
antenna and measure the wavelength A. (a)
Repeat, using a different inductance Le
and get A» Then the two wavelengths Fra. 109.—To measure capacity
are related as below. of an antenna.
Ar = 1.884 V (Li + La)Ca
1.884 V (Le + La)Co
where L, = antenna inductance in microhenries;
C, = antenna capacity in micro-microfarads.
i---
=>
a)
—
de
ll
Eliminating C, between these two equations we get
= Dyd2? a, Ded?
by, NE = re?
Problem 1-8. A wavemeter is being calibrated from a standard. At
resonance the capacity of the standard is 400 mmfd., the capacity of the
the other meter is 500 mfd. What is the ratio of their inductances? If the
inductance of the standard is 300 microhenrys, what is the inductance of the
other wavemeter? At what frequency are they now set? What is the wave-
length?
Problem 2-8. A coil and condenser combination has a resistance of 8 ohms
at 300 meters. What is the width of the frequency band at the points below
162 PROPERTIES OF COILS AND CONDENSERS
and above resonance where the current is 0.707 of its resonance value? The
inductance is 250 microhenrys.
Problem 3-8. In an experiment to determine the resistance of a coil by
the change of total resistance method (Experiment 5), the current without
added resistance is 100 milliamperes and with 12 ohms added it is 60 milli-
amperes. The resistance of the meter is 5 ohms. What is the resistance of
the coil and condenser in series?
Problem 4-8. A certain coil-condenser (LC) tunes to a frequency, fu.
The condenser is changed by adding another to it so that the value is C».
f. VG
The circuit now tunes to a frequency, fz. Prove that — = —=.
fi V Co
Problem 5-8. Using the formula in Problem 4, what must be done to
the capacity to make a circuit, tune to twice the frequency, half the frequency,
double the wavelength, one-half the wavelength?
Problem 6-8. A coil-condenser combination tunes to 450 ke. when the
condenser is 600 mmfd. When an unknown condenser is placed in series with
the 600-mmfd. capacity the circuit tunes to 600 ke. What is the unknown
capacity?
Problem 7-8. An antenna tunes to 30@ meters when 100 microhenrys
are in series with it, and 400 meters when 300 microhenrys are in series.
What is the inductance of the antenna? Remembering that the two induc-
tances, L; or Le, and the inductance of the antenna La, are in series, and can
be added to get the total inductance, what is the capacity of the antenna?
What is its natural wavelength?
139. Typical receiving circuits.—The first essential of all re-
ceivers is an antenna to collect energy from the distant station.
This can be a single wire stretched out in the open. For broadcast-
frequency receivers under average conditions, it should be about
60 feet long. In order that a receiver may be transferred to other
antennas without the difficulties due to changes in tuning, some
engineers have eliminated the antenna from the tuning of the set,
and use it only as a collector. It is usually loosely coupled to the
receiver, so loosely, in fact, that very little energy is transferred at
the lower frequencies where the coupling is weakest. It is here
that tuning the antenna to series resonance with the incoming
signals helps considerably toward boosting their strength, and at
times the practice of series tuning the antenna has been em-
ployed. Because of the distributed capacity of the coil and the
minimum capacity of the condenser as well as the capacity of
apparatus attached to the coil-condenser combination the lowest
TYPICAL RECEIVING CIRCUITS 163
wavelength that can be attained without changing the circuit is
limited. The ratio of longest to shortest wavelength received on
broadcast tuners is about 3 to 1, that is, from 200 to 600 meters.
Because the wavelength varies as the square root of the capacity,
the capacity range of the condenser must be nine to one. If the
maximum capacity of the condenser is 500 mmfd., a maximum of
Wave Length—Meters
50__100 150 200 = 250 300=—«350—«400.—«450 «500 +550 ~—«600
500
450 |
Engineering Department
Hammarlund Mfg. Co. : he lia
400 lon &
SS
2 350 a
; ¥
a5 Ss)
1 300 [ i ee > iy
2 S
.3) S
= S)
= 250 C iy @
2 tx
° By) & «
= 200 =i alae al i RS RS ‘
9 » ANS
= 150 — — — DM
100 se
00
50
0
Fia. 110.—Curves showing the amount of capacity and inductance to tune to
a given wave length.
50 mmfd. can be across the inductance when the condenser is
turned to zero degrees and still cover the required frequency range.
This capacity is made up of the coil capacity, minimum capacity
of the tuning condenser, leads, etc.
The resistance of the coil has an important bearing upon selec-
tivity and to some extent upon the sensitivity of the receiver. As
we shall see later it has an important bearing upon the fidelity,
or quality, of signals as they emerge from the loud speaker. The
164 PROPERTIES OF COILS AND CONDENSERS
resistance of modern variable condensers may be neglected. ‘The
resistance introduced into coils and accessory circuits by large
nearby metallic masses may be considerable, and therefore coils, if
shielded, must be kept at a respectable distance from the metallic
shielding material.
Modern receivers usually possess from two to four stages of
amplification at radio frequencies, and from one to three stages of
audio-frequency amplification. In general the more radio stages
Turns per Inch
10 20 30 40 50 60
Inductance in Micro—Henries
Fig. 111.—Curves showing the inductance of a coil as a function of length of
winding and turns per inch.
the more selective and sensitive the set will be, but the number of
tubes is no criterion as to the set’s ability to get weak distant dis-
tant stations through strong local interference. A well engineered
receiver of four tubes may be much better than a poorly designed
receiver of eight tubes. The difference lies not in the number of
tubes but in the design and subsequent engineering.
The manner in which L, C, \, and coil dimensions are related
is plotted in Fig. 110 and Fig. 111.
CHAPTER IX
THE VACUUM TUBE
THE most important single device known to radio science is the
vacuum tube. Although it is true that some receivers exist which
use no tubes at all, and that those which are within a very short
distance of broadcasting stations and which use only head phones
can get along with a coil, a condenser, and a crystal detector, far
the greater number of receivers in this country use tubes, some of
them one, some two, many as high as twelve or more.
140. The construction of the vacuum tube.—As we know the
tube today it consists of a glass wall within which are three metallic
parts known as the elements. In the center is the filament which
may be made of tungsten, carbon, tungsten covered with thorium,
platinum or nickel coated with oxides of barium, strontium,
caesium, and other chemical elements. Next to the filament is
the grid, an open mesh of molybdenum (frequently) wire screen;
finally is the plate which is a sheet or screen of metal, often of
nickel. Some tubes have only two elements, the filament and the
plate; many have an additional grid; the mechanical construction
differs according to type of tube, its use, and its manufacturer.
After the various elements are placed within the tube, the glass
wall is attached to a pump and the gas is removed. During the
pumping process the glass wall is heated in an electric oven to
drive out the gas from it, and later the elements are heated by
means of an “ induction furnace ”’ so that various gases bound up
in these metals may be pumped out. The modern tube is a high
vacuum tube; early types were poorly pumped and were really
gaseous tubes, tubes which would be rejected by modern testing
methods. When the pumping or “ exhaustion” process is com-
plete the glass wall and its contents are sealed. Then the tube
165
166 THE VACUUM TUBE
goes through several electrical tests and inspections before it can
be labeled, packed, shipped and again unpacked and sold for use.
141. The purpose of the filament.—In Chapter I of this book
we discussed the electron, that elementary constituent of matter
which carries electricity. Little has been said about the electron
in subsequent chapters; now it enters again and assumes an
important role. The filament is the heart of the vacuum tube;
the electrons which rush about in this filament are the life blood.
When the filament is dead—due to age or crossed wires—the elec-
trons no longer move in the proper manner; the tube is dead and
might as well be broken up. If a filament of tungsten is heated so
that an individual electron gets up a speed of 1 X 10° centimeters
per second (620 miles per second) it can break through the surface
tension of the filament. Since it is negatively charged it will be
attracted toward any positive body nearby.
142. The purpose of the plate-—When the electron is released
from the filament it goes shooting out into the void in which the
elements are situated. When it leaves the filament, it takes with
it a negative charge, and thereby leaves the filament positively
charged. If there is no body at a positive potential within the
bulb other than the filament, the electron will eventually find
its way back to the source whence it came. If, however, a
“plate ” is within the tube and is more positive than the filament,
the electron will be attracted to it. Even when the plate is at the
proper positive potential to attract many electrons, some go back
to the filament, and others congregate somewhere between the
filament and the plate and constitute what is called the “ space
charge.”
Every electron which hits the plate constitutes a minute electric
current and when enough of them arrive per second a measurable
current is attained. It is this current carried by the electrons from
the filament which constitutes the tube’s plate current which is
used in so many ways. The symbol for plate current is I; the
plate current is usually measured and expressed in milliamperes.
The source of the electrons is usually called the filament
although some modern tubes get their supply of electrons in another
manner, as indicated in Section 166. This filament is heated by
THE PURPOSE OF THE PLATE 167
a battery, called an A battery, or by a small step-down transformer
from the a.-c. lighting circuit. A battery inserted between part of
the filament system and the plate maintains the plate positive
with respect to the filament. It is called the B battery.
When the filament is heated to a proper temperature a copious
stream of electrons is emitted. Some of the electrons are attracted
to the positive part of the filament, that is, to the side of the fila-
ment, that is attached to the positive end of the A battery. If
the plate is insulated from the filament, a few electrons will get
through the fog called the space charge but if it is at a higher
potential than the filament it attracts many more electrons. It is
a.£,=B+6=51
c. E==B+0=45
Fig. 112.—If the B battery is connected Fig. 113.—Circuit for test-
as at (a) the voltage on the plate is 51 ing effect of plate voltage
volts; if as at (c) the voltage is 45. on emission.
usually so maintained by means of the B battery and in some cases
may be as high at 10,000 volts above the potential of the filament.
This B battery may be attached to the filament in several ways.
Its negative end may be connected to the negative end of the A
battery or to the positive end of the A battery. It is standard
practice in the telephone plant to connect A plus and B minus
together; in other places it is common practice to connect the two
negative leads together. The most negative part of the filament
in the case of d.-c. tubes, or the center of the filament in the case
of tubes run from a.c., is considered as the point to which all
other voltages are referred. (See Fig. 112.)
168 THE VACUUM TUBE
Experiment 1-9. Effect of plate voltage on a two-element tube.—Con-
nect the grid and plate of an ordinary receiving tube together and connect into
a circuit as shown in Fig. 113. Use a plate-current meter reading about 5 milli-
amperes. Light the filament and read the plate current as the plate is con-
nected to the negative end of the battery and then to plus 2, 4, and 6 volts
by connecting it to the first, second, or third storage cell in the battery. Con-
nect the negative terminal of a 4.5-volt C battery to the positive end of the
A battery and the positive terminal of the C battery to the plate. Read the
plate current. The plate is now plus 10.5 volts above the potential of the
negative part of the filament. Explain why.
143. Effect of filament voltage.—The experiment above shows
that the effect of increasing the positive potential of the plate is
increase the flow of electrons. The fila-
ment temperature, too, has an impor-
tant effect upon the flow of electrons.
The hotter the filament the more elec-
trons per second will be released into
the space surrounding the heated ele-
ment. If, however, the voltage on the
plate is low, there will soon be reached
a definite plate current which cannot
be exceeded no matter how hot the
filament becomes. In other words the
plate is taking all the electrons it can
Fig. 114.—Saturation curves. 8¢t through the space charge. It is
true that more electrons leave the fila-
ment at higher temperatures but they simply add to the space
charge or return to the filament. If the plate battery voltage
is increased, a greater plate current will flow, but again a
point will be reached where passing more current through the
filament ceases to increase the plate current. Typical saturation
curves for a 201 A type of tube are shown in Fig. 114. Whenever
the space charge (negative) is more effective in repelling electrons
than the plate (positive) is in attracting them a flattening plate
current takes place.
Ip Milliamperes
Experiment 2-9. Effect of filament voltage: Three-element tube.—A study
of many of the tube’s characteristics may be made with a set-up of apparatus
like that in Fig. 115 which consists simply of a board upon which are connected
SATURATION CURRENT 169
several Fahnestock clips to which may be attached meters and batteries of the
proper potential. A good voltmeter is a two-range Weston Model 506 reading
on the low scale up to 7.5 volts and on the upper to 150 volts. Jewell pattern
77isasimilar meter. These will read the ordinary ranges of filament and plate
voltages. A plate current meter may be any milliammeter reading from 5 milli-
amperes upward.
Fig. 115.—An experimental set-up for measuring tube characteristics.
Connect up a tube as shown in Fig. 116 and after reading the plate voltage,
place the meter across the filament. Use at the start about 22.5 volts of
B battery. Turn on the rheostat slowly and read the filament voltage and
plate current. If either meter should read backwards, reverse it. Plot as in
Fig. 114 the relation between Hy (filament
volts) and J, (plate current). Increase the
plate voltage and repeat.
144. Saturation current——With a
given filament voltage (which produces
a certain filament temperature) more
and more electrons will be drawn to the
plate as the voltage of the latter is in-
creased—up to a certain point. But
beyond this point additional plate volt-
age has little effect on plate current Ja}aafa}a}a}t|
and the plate current curve flattens out. Fig. 116.—Circuit for appara-
All of the electrons emitted by the fila- tus of Fig. 115.
ment are being taken by the plate and
increasing the plate voltage has no effect upon the number of
170 THE VACUUM TUBE
electrons emitted. Increasing the filament temperature produces
an additional supply of electrons, and the plate current will again
increase.
Experiment 3-9. Effect of plate voltage: Three-element tube.—Connect
up the apparatus used in Experiment 2 as shown in Fig. 116. Set the filament
voltage at some fixed value and take data showing the effect upon plate cur-
rent of varying the plate voltage. Increase the filament voltage and repeat.
Plot the data in a manner similar to that in Fig. 114. Remembering that 6.28
>< 1018 electrons per second flowing past a certain point in a circuit constitutes
an electric current of an ampere, calculate the number of electrons that arrive
at the plate per second for several values of filament and plate voltage.
The experiments and curves above show
1. The relation between plate and filament voltage and plate
current.
2. The saturation effect at low filament and plate voltages.
Saturation due to insufficient plate voltage is known as filament
saturation; that due to insufficient electron supply is called plate
saturation.
3. The fact that little is to be gained by increasing the filament
voltage above the rated value.
4. The curve connecting plate current (J,) and filament voltage
(E;) is not a straight line.
This shows that Ohm’s law is not being followed; the law in
fact is much more complicated. The plate current is zero at zero
filament voltage, and as the latter is increased the plate current
begins to rise too, but not in a straight line. Soon, however, the
negative space charge built up by the electrons which do not get
to the plate prevents any more electrons getting to the plate. The
plate current then is limited, and may be increased only by increas-
ing the plate voltage so that it is again more positive than the
space charge is negative. Various means are used to overcome this
space charge which shall be discussed later.
145. The purpose of the grid.—The third element, for which
DeForest is famous, is the grid, the mesh of wires between the
filament and plate. It has several important uses. It may be used
to neutralize the space charge so that greater plate current may
flow with a given filament temperature and given plate voltage.
CHARACTERISTIC CURVES 171
Suppose the grid is made positive with respect to the source of
the electrons. Since it is physically nearer the filament than is the
plate, a small positive potential will have the same effect as a large
positive potential on the plate. A positive potential near the
filament accelerates the escape of electrons, and prevents the
building up of a high negative space charge and then the plate has
a greater ability to attract the carriers of electric current.
Suppose, however, we make the grid negative. Owing to its
relatively close position with respect to the source of electrons, a
small negative voltage on it will counteract a large positive voltage
on the plate. So, with a small negative voltage we can prevent any
electrons from getting to the plate, or by varying this grid voltage
we can regulate in any desired way the number of electrons that
reach the plate, and thereby control the plate current. Since there
is no time lag in the flow of electrons, the grid voltages take instan-
taneous effect upon the plate current. The grid, then, is a control
electrode. The relation between the effects upon plate current of
the grid voltage compared to the plate voltage constitutes an
important tube ‘ constant,” the amplification factor.
146. Characteristic curves.—In the average receiving tube
there are three electrodes. The filament has already been men-
tioned, and the manner in which its temperature affects the plate
current has been tested. The plate and the effect of its voltage
on plate current have been qualitatively mentioned; so has the
effect of grid voltage. Under ordinary conditions the filament is
operated ‘‘ saturated,” that is, at such voltage that there is little
use in raising it further. Its voltage is then fixed; it is not varied.
If the filament voltage is considered as fixed, we still have the
plate current depending upon two variable quantities, the grid
and the plate voltage (H, and H,). The manner in which these
variables affect the plate current controls the characteristics of the
tube. When plotted in graphs, they are called characteristic
curves.
Experiment 4-9. Effect of grid bias upon plate current.—Set up the
apparatus as shown in Fig. 117, using in succession several of the common
types of tubes. Set the filament at the proper voltage. Fix a small voltage,
say 22.5 volts, on the plate of the tube and take down data showing how the
172 THE VACUUM TUBE”)
plate current changes as the grid bias is varied from a point where the plate
current is zero to a
grid voltage of about
positive 10 volts. The
D. P. D. T. switch in
the grid circuit
makes possible chang-
ing the polarity of the
grid without chang-
ing the meter (£,)
connection, Then
raise the plate volt-
age and repeat. Plot
these data like those
lanai + ALE obi
Fig. 117.—Complete apparatus for measuring charac- 147. Grid volt-
teristics. The D. P. D. T. switch reverses the grid age—plate current
voltage. curves. — Several
interesting and im-
portant facts may be discovered by looking at such curves which
we shall call the
E,—I, curves. At
large negative grid
voltages there is
little or no cur-
rent in the plate
circuit. As this
negative voltage
is decreased, some
electrons get past
the grid and
through the space
charge and to the
plate. The cur-
rent begins to
Plate Current Milliamperes
le
-60 =55 =50 5 “40 -35 -30 -25 “20 =15'-10) =5 9-0
flow, increases at Grid Voltage
a rather slow rate, Fig. 118.—A family of E,-I curves.
then morerapidly,
then in a steep and straight line, and finally, if the experiment is
GRID VOLTAGE—PLATE CURRENT CURVES 173
carried far enough, the curve flattens out. Increasing the plate po-
tential and again varying the grid potential produces a new curve
which is essentially parallel to the first, but moved to the left. In-
creasing the plate voltage again a like amount produces a new curve
displaced an equal distance to the left of the second line. Such a
graphic collection of data is known as a “ family ” of curves and
tells all we need to know of the effect of grid voltage upon plate
current. Grid voltages may be secured from a battery known
eee
[lial BSE ey Abe
ae 8
Fig. 119.—Grid current curve of a commonly used tube.
asa C Sey and the voltage itself is een called a “*C”
or grid “‘ bias.”’
If we place a meter in the grid circuit at the same time the
plate current is measured, we shall see that a very small grid cur-
rent is taken at positive grid potentials. This current in ordinary
practice is very small, seldom over one-tenth of the plate current,
and in all amplifiers in which the minimum distortion is desired
the current in the grid circuit is kept as low as possible by making
the grid highly negative. The grid current curve for a typical
case is shown in Fig. 119.
174 THE VACUUM TUBE
148. The effect of plate voltage upon plate current.—To deter-
mine this effect we will resort to experiment.
Experiment 5-9. Set up the apparatus as in Experiment 4. Set the grid
voltage at some value, say minus 5 for an ordinary receiving tube, and note
down the plate current as the plate voltage is changed from 0 to perhaps
100 volts. Then change the grid voltage to minus 10 and repeat; then at
minus 15 and 20; 0, and plus 5 and 10, etc.
14
13
CX-220
Plate Current mA
[o>} ~N foo} wo
€
20 40 60 80 100 120. 140 160 180 200
Plate Voltage
Fig. 120.—Plate current—plate voltage curves.
These data may be taken from the E,—J, curves plotted in Experiment
4 by picking off the curves the proper values of current, and plate and grid
voltages.
149. Plate voltage—plate current curves.—Here again (Fig.
120) the curves which we shall call the J, curves are essentially
parallel over the straight parts. If the grid voltages chosen are in
equal steps, the plate current curves will be equal distances from
each other.
From characteristic curves of this type, we may calculate all
of the tube’s constants, and foretell nearly all of its properties
AMPLIFICATION FACTOR 175
when connected into a circuit with other apparatus whose electrical
constants we know.
150. Amplification factor.—For example we know that the grid
potential is relatively more important in controlling plate current
than is the plate voltage. Why? Because it is nearer the source of
on eae
80 100 120 140 160 180 200
Fig. 121.—Detailed #,-I curve showing how to calculate R,.
electrons. How much? We can tell from the #,—J, curves in
Fig. 121. Looking at the line marked EH, =— 22.5 only, we see
that at a plate voltage of 180 the plate current is 16 milliamperes
but that if H, is decreased to 140 volts the plate current, decreases
to 8.2 milliamperes—a change of 7.8 milliamperes for 40 volts or a
net change of 0.195 milliampere per volt. This is the slope of this
176 THE VACUUM TUBE
particular line and, as we shall see later, it gives us another impor-
tant tube constant without further calculation.
Now looking at the two curves marked HE, =— 12.5 and
E, = —22.5 at the points where they cross the 140-volt E, line,
we see that at this value of plate voltage the plate current is respec-
tively 14.0 and 8.2 milliamperes at these two values of grid voltage.
This means that changing the grid voltage by 10 volts causes a
change of 5.8 milliamperes in the plate current, a net change of
0.58 milliampere per volt. Dividing the change per volt caused by
E, variations, by the change per volt produced by H, variations
gives us the relative ability of the grid and plate potentials to
influence the plate current. Thus,
Ability of grid voltage to control plate current
Ability of plate voltage to control plate current
0.58 ma./volt
SS ee ee
0.195 ma./volt :
This ratio is defined as the amplification factor of the tube. It
is the ratio between the plate voltage change required to produce
a certain plate current change and the grid voltage change required
to produce the same change in plate current. The Greek letter
mu, w, is the symbol used in the literature for the amplification
factor of a tube. Thus
_ plate voltage change to produce a given plate current change
o gridvoltage change to produce the same plate current change
The amplification factor for a given tube does not vary much
under the conditions under which the tube is ordinarily used. It is
controlled largely by its mechanical construction, and the nearness
of the grid to the filament. A grid composed of many wires close to
the filament produces a high amplification factor; a tube with a
wide mesh and not so close to the filament produces a tube with a
low amplification factor.
The student should note that the amplification factor is not the
ratio between plate and grid voltages, but is the ratio between
EQUIVALENT TUBE CIRCUIT 177
changes in these voltages. It may be expressed in more mathe-
matical language as
P
dE,
= to produce a given dI,,
where the prefix “d”’ signifies “‘a change in” (a differential).
The amplification factor may be obtained from the E,-I 3
curves in Fig. 121 in a manner similar to that outlined above. It
is also equal to the change in J, produced by, say, 10 volts change
in H, divided by the change in J, produced by 10 volts change
in #,. Thus in Fig. 120 changing E, from — 22.5 to — 12.5 (with
FE’, = 140) produces a variation of 5.8 milliamperes while along the
E, =— 22.5 line a change of E, from 140 to 150 volts produces a
variation of 1.95 ma.
5.8
Then yp = 195 3.0 as before.
151. The meaning of the amplification factor.—If the amplifica-
tion factor of a tube is 3, for example, adding 30 volts to the plate
will increase the plate current a certain amount. Adding only
10 volts (positive) to the grid will produce the same plate current
change. That is, adding 10 volts negative to the grid will bring
the plate current back to its value before the plate voltage had
been increased. In other words any voltage placed on the grid of
such a tube has the same effect as a voltage in the plate circuit
multiplied—or amplified—by the y» of the tube. A voltage H, on
the grid becomes equal to »H, when it gets to the plate circuit.
152. Equivalent tube circuit.—Since a change in plate voltage
may be replaced by a smaller change in grid voltage multiplied
by the u of the tube, we may replace the entire tube by a fictitious
generator whose voltage is »#, and whose internal resistance is
equal to the resistance of the tube. In fact in all problems the tube
is so considered, and is indicated symbolically as in Fig. 122.
Problem 1-9. With a 20-volt bias the plate current of a tube under a
given value of H, is 55 ma., and when the bias is increased to 30 volts the plate
current is reduced to 28 ma. If, however, at this value of grid bias (—30)
178 THE VACUUM TUBE
the plate voltage is increased from 180 to 210 volts the plate current comes
back to its original value, 55 ma. What is the amplification factor of the tube?
Problem 2-9. The amplification of a tube is 8 and when the grid is — 3
volts the plate current is 3 ma. If the bias is reduced to zero the current in-
creases to 8ma. Both of these current values were read when the plate voltage
was 90 volts. How much would the plate voltage have to be reduced (at zero
grid bias) to bring back the current to its 3 ma. value?
Problem 3-9. Changing the plate voltage of a power tube from 100 to
300 volts changes the plate current from 10 to 55 ma. If the bias is zero at
the latter figure what must be done to it to reduce the current to its former
value if the amplification factor is 7.8?
uRe
Eg
Lee >| [at
Fig. 122.—The tube and its equivalent circuit; a voltage »Z, in series with R,
and the load.
153. D.-c. resistance of a tube.—Since a certain plate current
flows under the pressure of a certain plate voltage, the ratio
oe
I,
in which E, = the d.-c. voltage on the plate;
I, = the d.-c. current in the plate circuit,
gives the d.-c. resistance of the space between the filament and
the plate. The power used up by the electronic current may be
found by multiplying the plate voltage by the plate current, or
9
EB?
Pia iE. OF e OV i ei.
This power is the rate at which the kinetic energy possessed by the
moving electrons is given up to the plate. When the electron
leaves the filament it is attracted toward the plate, increasing in
INTERNAL RESISTANCE OF THE TUBE 179
speed as it gets closer and closer to the positive potential which is
the attracting force. When the electron hits the plate its kinetic
energy due to its motion is given up. Ina transmitting tube the
number of electrons that arrive per second may be so high that the
plate becomes red- or white-hot.
154. Internal resistance of the tube.—This d.-c. resistance is
not what is popularly known as the ‘‘ impedance ” of the tube, or
more properly called its ‘‘ plate resistance” or “‘ differential or
internal resistance.” The latter is the ratio between a change in
plate voltage and the change in plate current produced by this
change in plate voltage. It is the resistance offered to the flow of
a.-c. currents in the plate circuit and is not the same resistance as
is offered to the flow of d.-c. current from the battery. Then
change in plate voltage dE,
OL d
? change in plate current dl,
For example changing the plate voltage from 180 to 140 volts
(Fig. 121) produced a plate current change of 7.8 milliamperes
(0.0078 ampere)
dE, 180 — 140 40
= — = ——— = — = §200 ohms (approx.).
Ry = 47, ~ 016 — 0082 .0078 epPE!
Problem 4-9. Make a table showing the d.-c. resistance of various tubes
in general use at the conditions they ordinarily work, that is, a 201-A tube
at HE, = 90 volts and E, = —4.5,a 171 with H, = 180 and E, =— 40.5. Use
values of plate current in the tube chart. Compare the d.-c. resistance with
the a.-c. resistance.
Problem 5-9. The plate current of a tube is 4.5 ma. when H, = 90 volts,
and is equal to 0.9 ma. when H, = 40 volts. What is the plate resistance?
Problem 6-9. The plate resistance of a tube is 12,000 ohms. At Hy =
140, J, = 14.5 ma. What is Ip when Hy = 100 volts?
Problem 7-9. Calculate the d.-c. resistance of the tube under the two
conditions of EZ, and J, in Problem 6, and the power used in heating the plate.
The internal resistance changes with plate and grid voltages
and so the conditions of both must be considered when the resist-
ance is mentioned. Thus the UX 171 has an internal resistance
of 2000 ohms when the plate voltage is 180 volts, and the grid
180 | THE VACUUM TUBE
voltage is 40.5 volts negative. Its internal resistance differs if
either H, or H, is varied.
The student should note that R, is not the ratio between a
plate voltage and a plate current but is a ratio between changes in
both plate current and plate voltage. Thus a 201—A tube at a
plate voltage of 90 has a plate current of 2.6 ma. The ratio
90
.0025
13,000 ohms under these conditions.
155. Mutual conductance of a tube.—There is one more impor-
tant tube constant, the mutual conductance. This is the factor
which tells us how much plate current change is caused by a given
grid voltage change. Thus
= 36,000 ohms is the d.-c. resistance; FR, is equal to about
change in plate current dl»
Gm = =
"change in grid voltage dz,
Thus if a change of one grid volt produces a change of plate
current of 1 milliampere, the mutual conductance
Pele L0 =
Gm
1
= 1 X 10~-% mho or 1000 micromhos.
The mutual conductance is defined, too, by the ratio between the
amplification factor and the plate resistance. Thus
dl mM H dH
Cs Dia ag A &: Pp = P
inl e Re because u dE, and R, as
dB
Therefore sei a ans = uy
Tes dE, dE,
dl,
Problem 8-9. A tube has a plate current of 7.75 ma. at zero grid bias
and 3.8 ma. at Hy =—4. What is the mutual conductance?
Problem 9-9. The mutual conductance of a tube is 800 micromhos. What
change in plate current is produced by a 1 volt change on the grid?
Problem 10-9. The mutual conductance of a tube is 775 micromhos.
The plate current at E, =—6is2ma. What is the plate current at E, =—2?
SLOPES OF CHARACTERISTIC CURVES 181
156. Importance of mutual conductance.—The tube is ordi-
narily so worked that small variations in input (grid circuit)
a.-c. voltage produce variations in output (plate circuit) a.-c. cur-
rents, and it is important that the mutual conductances of a tube
shall be high. ‘Tubes are in general use which have amplification
factors as low as 3 and as high as 30, the plate resistance ranging
from 2000 to 100,000 ohms. When still greater amplification
constants are desired—we are speaking of small receiving tubes
only—it is necessary to change the construction of the tube. It is
here that the screen-grid, or four-element tube, arrives on the
scene. It is described in Section 170.
When one is considering two tubes of the same type, say two
201—A tubes, the one with the higher mutual conductance is the
better, but rather large differences in mutual conductance must
occur before any difference in the operation of a circuit in which
the two tubes are to be used will be noted. If one is to compare
a CX-301—-A with a CX—220 he will see that the mutual conduc-
tance of the latter is less than of the former, and yet the CX-301-A
is capable of furnishing only about one-half as much undistorted
power to a loud speaker as the latter. In comparing tubes of the
same type which are to be used for the same purpose the mutual
conductance is the best single factor—but it must not be worked
too hard.
157. Slopes of characteristic curves as tube constants.—Since
the plate resistance of the tube is defined as
R. = dE,
a dl,
and the mutual conductance as
dl
Gn — 2
dE,
these values may be taken directly from the curves showing the
relation between plate voltage and plate current and between grid
voltage and plate current; the reciprocal of the plate resistance is
i
_the slope or steepness of the #,—J, line, that is, pees slope of the
Dp
182
THE VACUUM TUBE
E,-I, curve; the mutual conductance
is the slope of the H,-I, line. When
changing the grid voltage produces no
change in plate current (saturation)
the mutual conductance is zero, that
is, the H,-I, curve no longer has any
slope or “steepness”? and it flattens
out, as for high positive values of grid
bias. On the other hand, when the
E,-I, curve flattens out the plate
resistance becomes infinite—and so the
Fig. 123.—Note how Rp is
obtained. It is not Hp, di-
vided by Jy, which gives the
d.-c. resistance.
plate resistance, mutual
conductance, and am-
plification factor. Plot
these values against grid
voltage for one curve
and against plate voltage
for another curve. Such
data for a typical tube
are shown in Fig. 124.
158. ‘‘ Lumped ”
voltage on a tube.
— An expression in
England for the volt-
age on the plate of
a tube is very use-
ful. It involves the
expression,
E = E, + wE,,
steeper the H,-I, curve the lower the
plate resistance. Care should be taken
that the plate resistance is obtained
properly from the #,-/, curve. Figure
123 shows the correct and incorrect
methods.
Experiment 5-9. From the curves plotted
in Experiments 4 and 5 measure the slopes at
various values of Hp and H, and calculate the
16 000
14 000
12 000
8 6 4 2 0
Eg (Negative)
Fig. 124.—Effect of grid voltage on tube charac-
teristics.
“LUMPED” VOLTAGE ON A TUBE 183
where EH = the effective or “lumped ” voltage on the plate;
E, = the plate voltage due to the B battery;
E, = the voltage on the grid, due a C bias :
u = the amplification factor of the tube.
For example we know that adding a negative C bias to a tube’s
grid reduces the plate current. The same value of plate current
may be attained without the C bias by reducing the plate voltage.
Then for every value of plate current there is a combination of
grid bias and plate voltage that will produce it, the actual values
being given by the above expression.
If, then, we plot the single curve showing the plate current at
various plate voltages but with the grid at zero bias, we can easily
calculate what the plate current will be under some other condi-
tion of B and C voltage. Thus with zero bias and 90 volts on the
plate, the plate current may be 10 ma. That is,
E=90+ 4X0 = 90
Now suppose wp =8 and EL, = —3
E = 90+ 8 X (— 3) = 90 — 24 = 66.
To get J,, it is only necessary to look at our H,-I, curve for
E, = 0 and find what J, is when EH, = 66. This is the value of
current desired, and is one point on the new curve E, = —3.
Then assume some other value of E, and find the new I, and place
a mark on the graph paper for this point which is the second for
the ZH, =— 3 curve. Other points may be obtained for other
values of EZ, and FE, and thus any number of curves drawn which
will be parallel to the first or #, = 0 curve.
Problem 11-9. The ,» of a tube is 3; what is the lumped or effective
plate voltage when H, = 180 and EH, = — 40?
Problem 12-9. The plate voltage HZ, on a tube is 120 but the plate cur-
rent corresponds to a plate voltage of 40 at zero grid bias. If the value of
E, is 10, what is the u of the tube? Do you see a simple way of measuring the
amplification factor of a tube by this method?
Problem 13-9. If the plate current of a tube is 2 ma. when EH, = 90 and
E, =—4.5 and » = 8, at what value of Z, will it be equal to 2 ma. when
Oren ae ,
E, = 0?
184 THE VACUUM TUBE
159. Measurements of vacuum tube constants.—The various
factors which define all of a tube’s characteristics—y, R,, and Gm—
are known as the tube constants. It is very important that means
be handy for measuring these constants at various values of plate
and grid voltage so that a full knowledge of a tube’s character-
istics may be had.
Such calculations may be made by means of the characteristic
curves. This is slow work, however, and the experimenter or radio
service man has no time for such processes. A simpler method
will be described.
(a) To measure the plate resistance of a tube.
This is the ratio between the change in plate voltage and the
change in plate current: dH,/dI,. Set the grid bias at the value
desired (for example — 4.5 for a 201—A tube) and choose the plate
voltage at which the tube will probably operate (that is, 90 volts
for a 201—A, 135 for a 112, etc.). Then fix the plate voltage at a
value somewhat higher than this and read the plate current.
Reduce the plate voltage to a value lower than that at which the
tube is to be used and read the plate current. Suppose we are to
test a 201-A type tube. The proper values are 90 for H, and
— 45 for E,. Set the grid voltage at — 4.5 and the plate voltage
at 100 and then at 80. Set down the data as below, in which the
estimated currents are respectively 3 and 1 milliampere:
dE, 100 — 80 20
Oe” WS Tn
Ry, = = 10,000 ohms.
This means that at a grid bias of H, =— 4.5, the average plate
resistance between the values of H, = 80 and 100 volts is 10,000
ohms.
(b) To measure the mutual conductance.
This is the ratio between plate current change d/J, and the
grid voltage change dE, that produced it. Set the plate voltage
at the value at which the tube will operate, for example: 90 volts.
If the bias under operating conditions is to be 4.5 volts (CX-301-A)
set the grid at minus 5 and then minus 4 and read the plate cur-
BRIDGE METHODS OF DETERMINING TUBE FACTORS 185
rents. Set down the data as follows, assuming the plate currents
are respectively 3 and 2 milliamperes:
_ di, 003 — 002.001
= iE, = nae = ia = .001 mho = 1000 micromhos.
Gm
(c) To measure the amplification constant.
The value of the amplification constant may be calculated
directly from the results obtained in (a) and (b) above. Since
Gin = i
R,
or [Ui Gin <4 sie
it is only necessary to multiply the plate resistance by the mutual
conductance. Thus if
R, = 10,000 ohms
Gm = 1000 X 10-® mhos
1000 X 10-® X 10,000 = 10.
I
ML
The amplification constant may be determined in the following
manner, which makes it unnecessary to determine first the mutual
conductance and the plate resistance.
Set the plate voltage at a certain value, say H,,, read the plate
current, I,,; change the plate voltage to H,,, note plate current
I,,._ Then bring the current J,, back to its original value J,, by
varying the grid voltage. The ratio of the corresponding plate
and grid voltage changes is the amplification constant, as indicated
in Section 150. That is
Ey, fe En,
1 eae
lip Nay
160. Bridge methods of determining the tube factors.—A num-
ber of meters have been devised to measure the three tube constants
by the methods outlined above—meters which read them directly
and in a very simple manner. Other methods of obtaining the
tube constants are in common use in laboratories. These methods
involve balancing out one voltage by another.
186 THE VACUUM TUBE
For example let us consider the circuit in Fig. 125. An a.-e.
current flowing through R; and 2 in series sets up two voltages
across them. The voltage Hi= I X Ri becomes EH, and is impressed
across the grid-filament input of the tube where it is amplified by
the tube to appear in the plate circuit as « times H,. When the
pridge is balanced by varying Ri and Ry» no voltage is across the
telephones and accordingly no sound is heard. At balance
plhy =I Re
uk, = Re
_ ky
hae Ri
Fig. 125.—Bridge for determining Fig. 126.—Equivalent of Fig.
amplification factor. 125.
It is only necessary to adjust the ratio between the two resistances
until the sound in the telephones is balanced out when the amplifi-
cation factor is equal to the above expression. The same balance
conditions may be found by substituting a battery and a key for
the a.-c. voltage, and a plate current meter for the telephones.
Pressing the key will change the plate current unless the ratio of
Fz to fF is adjusted properly. When no plate current change
occurs
TO MEASURE THE PLATE RESISTANCE 187
In either of these bridges, Ri may be fixed at 10 ohms and Re
varied. Thus if Ri = 10 ohms, and at balance Re = 134 ohms,
the u of the tube is 134 + 10 or 13.4.
161. To measure the plate resistance.—A bridge circuit may
be set up to measure the plate resistance. There are several such
bridges; one of them is shown in Fig. 127. Let us consider
Fig. 128 which is equivalent to Fig. 127 (neglecting L) in which
R, represents the internal plate resistance of the tube. Current
yn v,
Fig. 127.—-A bridge for measur- Fig. 128.—The equivalent of
ing Rp. Fig. 127.
I from the alternator flows through R; and Re as well as through
R, and R. When no sound is heard in the telephones
pee
Ri Re
or
Ry
fey TK
If R = 10,000 ohms
Rz = 100 ohms
R, = 100 X*fi
188 THE VACUUM TUBE
The inductance L is useful in balancing out certain
capacity voltages. It is not necessary and its reactance
can be neglected in calculating the tube constant being
measured.
It is essential that small variations of voltage be used. Since
the E,-I, curve is not straight, its slope differs at different points
and is only a “ constant” over a limited part of the curve. Ifa
large variation, dH,, is used to measure the plate resistance, or a
large grid variation, dH, when the mutual conductance is measured
the values obtained will not be very accurate. Under normal condi-
tions the a.-c. voltages put on the grid of a tube of the 201—A type
is seldom over 3 volts, and so it is absurd to measure the various
constants by varying the grid more than this amount. Since the
amplification factor of this tube is about 8, a grid voltage of 3
corresponds to a plate voltage variation of about 24, and so changes
in plate voltage, dH, greater than this value should not be used
when obtaining tube constants.
162. An a.-c. tube tester.—A simple tester which the diagram
in Fig. 129 describes pictorially will be useful for quick tests to
determine whether or not a
tube should be thrown away.
It comprises an a.-c. trans-
former which provides the
proper filament voltages for
standard tubes, and a resistor
divided into two parts through
which the plate current flows.
Across this resistance appears
Fig. 129.—An a.-c. tube-tester. a voltage which may be used
as C bias. The operation
of the tester is as follows: it is plugged into a lamp socket
(a.-c. of course), the tube is inserted, and the reading of plate
current noted. Then part of the grid bias resistor is shorted,
thereby changing the plate current and the C bias. The ratios
between the corresponding plate current changes and the grid
bias changes give an indication of the mutual conductance of the
tube. In actual tests it would determine the mutual conductance
TYPES OF TUBE FILAMENTS 189
of several types of tubes within 80 per cent or better of the value
as measured upon an accurate bridge.
Suppose for example that the plate current is 0.001 ampere
when the bias resistor is 4000 ohms and 0.003 ampere (3 milli-
amperes) when the bias resistance is reduced to 500 ohms. We
set the data down as
seed ie (0021-001 002g 002
eal wer 00 < 4000 — .003 52500) | 40 1h as
= .0008
= 800 micromhos.
163. Types of tube filaments.—The filament of the vacuum
tube has undergone many mutations since the invention of the
tube. What is desired, of course, is a copious emitter of electrons,
that is, one which gives off many electrons at a low filament
temperature and with expenditure of little current from the heating
battery. Tubes which burn with a dull glow—Western Electric
tubes, the WD 12, the 112-A of R. C. A., and Cunningham, the
rectifier tubes such as the UX 280, have filaments which burn at
a comparatively low temperature. They are very efficient because
they are coated with rare elements which emit many electrons
even though the temperature is not high.
A recent type of filament is the thoriated tungsten filament. It
is made of tungsten which is impregnated with atomic thorium.
When the filament is heated the thorium gives off the electrons,
and as the supply on the surface of the wire is exhausted a new
supply comes from the interior of the filament. If, due to an acci-
dental overload of filament voltage, the tube seems not to have the
required plate current, it is probable that the balance between
the rate at which the surface electrons are used up and the rate at
which they come from the interior of the filament has been upset.
The tube may then be reactivated, as explained in the following
section.
The thoriated filament is very efficient, requiring much less
power from the filament battery than any other kind of filament
for a given amount of power output. The filament in the 199 type
190 THE VACUUM TUBE
of tube is finer than the human hair, and requires only 60 milli-
amperes to give sufficient emission—which should be compared
with the 201 type of tube used not so long ago. It then required
1 ampere at 5 volts to heat it sufficiently to give off a rather
meager supply of electrons.
164. Reactivating thoriated filaments.—The older types of
tube usually evidence the end of their useful life by burning out.
The thoriated filament, on the other hand, has a practically con-
stant output until the thorium is exhausted and then the output
of the tube falls off rapidly although the filament still lights and
looks perfectly normal. The test for such cases is to connect grid
and plate together, apply a certain voltage to these common ele-
ments and the filament, read the current, and if it falls below a
certain value “ reactivate.”
Machines for reactivating thoriated filaments are on the market
and diagrams of connections for such devices and directions may
be found in the Cunningham Tube Data Book, 1927. No other
type of filament can be reactivated.
165. Alternating-current tubes.—Tubes which could be heated
by a.c. have long been sought by all workers in radio fields because
of the greater simplicity of operation. A transformer attached to
an a.-c. circuit is much less cumbersome than a storage battery
which must be periodically charged. There are several reasons
why we cannot run the ordinary type of tube from a.c. When
ordinary battery-operated tubes are lighted by a.c. an objectionable
hum results. Suppose we lighted the filament from a.c. The
voltage along the filament is continually changing, part of the time
one side is positive with respect to the other, and part of the time
it is negative. There is a continual heating and cooling going on
which cannot help but transmit some of its variations to the plate
circuit.
If a centertapped resistor is placed across the filament and the
plate and grid circuits are attached to the center of the filament as
in Fig. 176, the hum emanating from the plate circuit is less, but
the filament is too light and has too little thermal inertia to with-
stand the continual heating and cooling without transferring some
of its variations to the plate circuit. The hum is still too great.
HEATER TYPES OF TUBES 191
Suppose, however, we have a very heavy filament with a low
voltage across it. It has a high thermal inertia and it is possible
to get a good balance between the electromagnetic and electro-
static fields at the value of plate current desired by introducing the
grid and plate circuits to the center of the filament by means of a
centertapped transformer winding, or a resistor. Such is the 226
type of tube. This uses a very rugged oxide-coated filament which
consumes 1.05 amperes at a voltage of 1.5. The voltage drop across
the tube is low, its thermal inertia is high, it has a very low hum
output. It can be used as a radio-frequency amplifier and to some
extent as an audio-frequency amplifier.
166. Heater types of tubes.—The separate heater type of tube
has a metal cylinder which surrounds a filament heated by alter-
nating current. The cylinder is heated by conduction and convec-
tion from the filament proper from which it is electrically insulated.
(The Arcturus 15-volt heater tube has the heater and the cylinder
or emitter electrically connected.) The thermal inertia of the
cylinder and the insulating material is so great that fluctuations
in a.-c. voltage of the filament do not affect the plate current.
The hum appearing in the plate circuit of this tube is still less
than that occasioned by the use of the 226 type. It is a general
purpose tube and can be used in all positions in the receiver except
the final power stage. In the usual type of heater tube there are
five prongs, the additional one attached to the electron source,
known as the cathode, which is the element to which the grid and
plate circuits are connected. Less hum will be evidenced if the
heater (filament) is biased by the voltage determined by experi-
ment.
It is better to use a resistor with a variable center tap with the
226 type of tube in preference to a centertapped transformer wind-
ing. In this manner the best adjustment may be found after the
tube and associated apparatus have been assembled. The use of
a.-c. tubes either of the heavy filament (226) type or the separate
heater type (227) requires only slightly different apparatus com-
pared to d.-c. tubes. A typical a.-c. circuit is shown in Fig. 130.
167. Operating filaments in series——Under ordinary circum-
stances vacuum tubes are operated with their filaments in parallel,
THE VACUUM TUBE
1,000 Ohms
Brown
171
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OPERATING FILAMENTS IN SERIES 193
so that the total current taken is the sum of the filament currents of
the individual tubes. It is possible, however, to operate the tubes
with their filaments in series, and under some circumstances this is
preferable to the other arrangement.
It is more difficult to rectify and filter large currents at low
voltages than small currents at high voltages. If, however, it is
possible to get a well rectified and filtered source of current at say
250 ma. we can operate a radio receiver with ordinary tubes from
the a.-c. circuit by connecting the filaments in series.
00600
—30 +30
Fig. 131.—A series filament circuit in which the voltage drop across a tube’s
filament is used as another tube’s grid bias, etc.
Let us consider the circuit of Fig. 131 in which the filaments are
wired in series. In this case the same current flows through each
tube and the total voltage required is the sum of the voltage drops
across the tubes and the 10-ohm resistor. To operate a receiver
using five 5-volt tubes requires a source of current that will produce
250 ma. at a voltage of 5 X 5 or 25 volts plus whatever additional
C bias voltages may be necessary. The current enters one end of
the filament series and exits at the other. As we go along the
receiver there is a series of voltage drops, each filament as it gets
194 THE VACUUM TUBE
farther and farther away from the positive terminal becoming more
negative with respect to the preceding tubes. The drops in vol-
tage along the line may be utilized as C bias voltages for other
tubes.
For example suppose the final tube (No. 5) is a power tube
requiring a C bias of 10 volts. We note that the negative terminal
of the filament of tube 3 is 10 volts more negative than the negative
terminal of tube 5 because there is a 5-volt drop across the filament
of tube 4 and an additional drop of 5 volts across tube 3. We attach
the grid circuit to this point in the circuit. If tube 4 requires only
5 volts C bias we can attach its grid circuit to the same point in the
circuit because so far as tube 4 is concerned this point is only
5 volts negative. Ifthe detector tube 3 requires a C bias of positive
5 volts we can attach its grid circuit to the positive leg of its own
filament.
Suppose the first two tubes in the set do not require 5 volts
bias but some value less than this. All that is necessary is to insert
a resistor in the circuit ahead of that tube so that the negative drop
in voltage in this resistance will be utilized. Thus in Fig. 130 the
250 ma. through a resistance of 10 ohms will produce a drop of
2.5 volts. The grid circuit should be attached as shown.
The 250-ma. supply may be a rectifier such as is used to charge
storage batteries, or it may be a gaseous tube of the Raytheon
type, or a chemical rectifier or some other means of rectification.
The filtering is a big problem, and has limited the use of such a
system. It is fairly simple, however, to get a 60-ma. current which
may be used to light the filaments of 199-type tubes, and many
series filament receivers have been built utilizing this scheme of
connections.
168. Compensating for plate current flowing through filaments.
—It should be noted that the plate current of tube 5 to return
to the negative side of this tube’s filament must pass through the
filaments of all the other tubes. in a similar manner the plate
current of tube 4 must pass through the first three tube filaments.
The filament of tube 1 not only has 250 ma. (or 60 ma.) through it
but the plate currents of the other tubes as well. This may cause
serious overloading of the tube filament, and means must be taken
COMPENSATING FOR PLATE CURRENT 195
to avoid this. A resistance shunted across the filaments such that
the total current passing through the tube is only its normal value
is one way of avoiding trouble of this kind.
Fig. 132.—A series filament circuit in which the plate of the power tube is in
series with filaments of the other tubes.
The plate of the final tube in the receiver may be connected in
series with the filaments and resistors of the other tubes, as in
Fig. 132, or in parallel to this string of tubes as in Fig. 133. In
case the final tube takes 60 ma. of plate current its plate circuit
may be in series with the filaments of the previous tubes. If this
C Bias Resistor and
Resistor for obtaining a Voltage
drop for Plate Voltage
To reduce 350 Volts to Value 60 m.a.
needed for Series Filaments
<—
80 m.a
Fig. 133.—Series filament circuit in which the plate of the power tube is in
parallel with the filaments of the other tubes.
final tube takes only 50 ma. of plate current, a resistance placed
in parallel with the tube will increase the current to the required
amount. One system requires a high voltage at a low current;
the other requires a lower voltage at a higher current.
In all such systems the current must be well rectified and fil-
tered. Because the plate current of one tube may flow through the
filament of another tube, or through a resistor which is common to
196 THE VACUUM TUBE
some other tube’s grid or plate circuit, there is always danger of
regeneration and unwanted couplings. These may be reduced by
proper by-passing and filtering of all a.-c. circuits as described in
the following chapter.
169. Means of obtaining C bias in amplifier tubes.—In all
amplifiers of the present time it is desirable to maintain the grids
of the tubes at a negative potential with respect to the filament.
Such biasing keeps the plate current low, and if properly done will
situate the operating point on such a part of the characteristic
that minimum distortion due to overloading and curvature results.
EES:
60~
110V
2000
Ohms
Grid Return Lead s oe
Fig. 134.—Obtaining grid Fia. 135.—Obtaining grid bias by means
bias by means of filament- of plate current flowing through a resist-
current flowing through a ance.
resistance.
Bias voltages may be obtained from a C battery. The voltage
drop across a resistance in the tube’s own filament may be used as
bias, as is shown in Fig. 134. Remembering that all voltages in
the circuit are reckoned either from the negative side of the fila-
ment or from the center of the filament (or the cathode in the ease
of heater tubes) we see that the grid of this tube is negative
by the amount of the voltage drop in the resistance. Attaching
the grid “ return ” to this point gives it a 1.0-volt bias. Another
method involves using a resistance through which flows the plate
current of the tube (or several tubes). Thus in Fig. 135, which
represents the conventional power stage for the radio receiver, the
2000-ohm resistor is connected to the center tap of the filament
transformer. The plate current flows through this resistor, pro-
SCREEN-GRID TUBE 197
ducing a voltage drop ere. and since the end toward the B battery
is more negative than the end toward the filament, the grid return
lead of that tube may be connected to the same point as the nega-
tive B lead. The grid is then biased negatively by the voltage
drop in this resistor. If the plate current is 20 ma. the grid bias
will be equal to .020 X 2000 = 40 volts. If because of an increase
in plate voltage from any cause the plate current increases to 25 ma.
the grid bias increases to 50 volts—and such a biasing scheme pro-
vides some protection for the tube. When the plate voltage
changes, the grid bias changes too and tends to keep the plate cur-
rent within prescribed limits.
170. Screen-grid tube.—In many uses, notably as radio-fre-
quency amplifiers, the capacities which exist within the tube are
detrimental. There are three such capacities, the grid to filament,
grid to plate, and filament to plate. Because of the grid-plate
capacity a path exists between the input (grid circuit) and the
output (plate circuit) of the tube so that variations in the output
may affect the input. Because of the amplification factor of the
tube these variations may be amplified and repeated back into the
plate circuit and the performance repeated until the normal func-
tioning of the tube is seriously impaired. It would be an advantage
if the grid-plate capacity of tubes could be eliminated or reduced.
Because of the space charge—the cloud of negative electrons
between the filament and the plate—the current that can flow in
the plate circuit is limited, and if this space charge could be
eliminated, or at least reduced, the grid voltage would have a
much greater controlling effect on the plate current.
Both of these beneficial effects may be secured by the screen-
grid tube. In this tube, the grid-plate capacity has been reduced
from an average value of 6.0 mmfd. to about 0.02 mmfd. and an
amplification factor of several hundred is not difficult to attain
and at the same time the plate resistance of the tube does not be-
come prohibitively great.
The screen-grid tube consists of the usual elements and an
additional grid. The second grid is maintained at a positive poten-
tial with respect to the heater or filament and thereby does away
with the space charge as well as serving as a shield for the plate.
198 THE VACUUM TUBE
In Fig. 136 is an illustrative example of what the second grid
does to the tube. If any difference of potential exists between the
plates P and G, a capacity exists between them, and any a.-c.
voltage attached to the system will cause an a.-c. current to flow.
Now if a plate is placed between P and G and grounded, the cur-
rent as read by the current meter drops to zero, because this part
of the circuit has been shielded or protected from the a.-c. voltage.
The capacity between P and G, then, becomes zero.
In the tube P and G are the plate and control grid respec-
tively and the extra grid is the screening plate. It is impossible
G P to screen the plate from the grid completely
because some electrons must get to the plate
through the grid, or the tube would be worth-
less. The screen grid is fixed at some positive
potential lower than that of the plate. The
fineness of this grid and its position controls
wn the screening effect upon the plate. The
Fig. 136.—Capacity electrons from the filament proceed toward
current from grid to the screen grid at considerable speed and
plate is nullified by most of them go through it and are collected
grounded shield or : ee ‘
a ceN by the plate provided it is at a higher poten-
tial than the screen. Because of the inter-
position of the screen grid between the plate and the control grid,
the rate at which electrons go across the space is not controlled so
much by the plate voltage as it is by the voltages on the two grids.
In other words the plate current is more or less independent of the
plate voltage, and the plate resistance—which is the ratio between
I
\
|
|
{
!
|
|
=
’ dE
changes in plate voltage and changes in plate current (ee)
Pp
very high, 750,000 ohms for the d.-c. tube (UX 222) and 400,000
ohms for the a.-c. tube (UY 224).
The mutual conductance of the a.-c. tube is about 1000 and the
amplification factor is 400. It has a 2.5-volt heater type of
filament.
Because of the high voltage amplification, the control grid must
be protected from all other wires and circuits. It is connected
only to a cap on top of the tube, and in practice is connected to its
CHARACTERISTIC CURVES 199
proper circuit by a wire which is covered with a grounded shield.
The entire tube is also covered and thus all stray voltages are pre-
vented from getting to the control grid. High a.-c. voltages may
get to the screen grid and this circuit should be carefully filtered.
171. Characteristic curves of the screen-grid tube.—Some of
the characteristic curves of the a.-c. screen-grid tube are shown
in Fig. 137. It will be seen that changes of plate voltage have
Average Static Characteristics
re | Radiotron UY 224
; A Heater Volts =2.5 T,=Plate Current
z Screen Volts=+75 Icg= Screen Current
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2 50 100 150 200 250
10 eB Plate Voltage
Values to left of vertical line Subject] to Considerable
2.0 Subject to Considerable Variation Variation
Fig. 137.—Characteristic curves of UY 224. In the region to the left of the
80-volt line there may be as much or more current flowing away from the
plate as toward it, due to ‘‘secondary emission.”
little effect on plate current; that at low plate voltages the cur-
rent actually decreases instead of increases. At plate voltages
lower than the screen grid voltage electrons may come from the
plate and go to the screen grid, thereby causing the current to the
plate actually to decrease. This backward-flowing circuit is due
to “ secondary emission.”’ That is, an electron from the filament
may get through the screen grid, but at the plate it dislodges an
electron and then both are dragged back to the screen grid because
of its greater positive potential.
200 THE VACUUM TUBE
The sum of the currents taken by the screen grid and the plate
is almost constant, and is never very high. The average plate cur-
rent is about one milliampere for the d.-c. tube and about 4.0 ma.
for the a.-c. tube. Normal voltages are 135 to 180 on the plate
and 45 to 75 on the screen grid, and minus 3.0 on the control grid.
When used as an amplifier, the tube is operated on some por-
tion of the long almost-flat part of its plate current-plate voltage
curve. Although the curved parts of the characteristic present
interesting possibilities, little has been done toward utilizing them.
Early receivers utilizing screen-grid tubes suffered from many
troubles. Among others was the production of cross-modulation
or cross-talk by strong undesired signals. For example a receiver
might be tuned to a fairly low strength station. This meant that
the volume control, usually either grid or screen voltage, was
turned up high so that the screen-grid amplifiers were working
near their maximum sensitivity. These tubes had such a short
range of grid voltage over which they could work that a strong
signal would force the grid voltage to the point where the a.-c. plate
current would cut off. This resulted in severe distortion, evi-
denced by blurbs or gasps of undesired modulation crashing
through the desired program, or by hum entering from the power
supply.
Late in 1930 a new screen grid tube made its appearance and
became of considerable importance in the following year. This
was the variable-mu or exponential tube. It had a very long,
fairly flat characteristic at the bottom of its HJ, curve. In
other words strong negative voltages on the grid did not force the
plate current to zero, and such cross-modulation due to plate
current cutoff was prevented.
Because of the very long even characteristic, the tube was
nicely adapted for automatic volume control, for radio and inter-
mediate amplifiers, and for recording signal strength, or loud
speaker output or other measurements where large ranges of cur-
rent or voltage, etc., were to be measured.
172. Space charge grid.—If the grids in the tube are reversed,
i.e., the normal screening grid is used as the signal or control grid,
and the normal control grid—the one next to the filament—is made
MISTREATMENT OF TUBES 201
positive, the tube may be used as what is known as a space charge
grid tube. The advantage of a much lower plate resistance and of
a large amplification factor is thereby secured.
The screening effect of protecting the grid from potential varia-
tions in the plate circuit is eliminated, however, and the tube is no
longer good at radio frequencies. Because of the high internal
capacities in this connection, the space charge grid tube tends to
discriminate against high audio tones when used as an audio
amplifier.
173. Mistreatment of tubes.—Overloading the filament of a
tube with higher currents than they are rated for is one of the
surest methods of shortening the life of that tube. If the voltage
across a tube is 10 per cent greater than normal (thoriated fila-
ments) the life will be decreased by much more than this per-
centage. It is also true that the plate current should not be
allowed to run higher than normal. Many receiver manufacturers
in the past have used general purpose tubes without adequate C
bias with the result that the plate current might be doubled over
its normal value. This plate current must come from the fila-
ment, and flowing through the tube’s resistance amounts to a cer-
tain amount of power which must be dissipated. Thus if a tube
has a resistance of 20,000 ohms and a normal plate current of
2 ma., the power that must be dissipated on the plate is 0.08 watt,
but if the current is permitted to become as high as 4 ma., which
without C bias is a reasonable figure, the plate power becomes
0.32 watt—four times as great. This heat may not damage the
filament, but it is certain to do it no good, and the electrons from
the filament are being used at a much greater rate than is necessary
or desirable. Turning up the filament voltage or increasing the
plate current in order to improve the operation of a receiver is a
sure sign that the receiver is poorly designed or engineered, and
that a new set of tubes will be needed before long.
The vacuum tube is a delicate device and must be treated with
the care one expends upon a good watch. When it is mistreated,
its life is shortened.
CHAPTER X
THE TUBE AS AN AMPLIFIER
In the usual receiver, at least four of the tubes are used as
amplifiers; two of them amplifying the signals at radio frequencies,
two of them amplifying the voice or musical frequencies after the
fifth tube has demodulated the amplified radio-frequency wave.
How does the tube amplify?
174. The tube as an amplifier—Let us look at the curve in
Fig. 138 which shows in exaggerated form the relation between
dl
plate current and grid voltage. The slope of the curve, — is
g
the mutual conductance of the tube and tells how many milli-
amperes the plate current changes when the grid potential is
changed one volt. The curve shows that adjusting the C bias on
the grid of the tube to 5 volts, that is, placing a 5-volt battery
between the grid and the most negative part of the tube filament,
permits a plate current of 3 milliamperes to flow. If this bias is
decreased to 4 volts, the plate current increases to 3.6 ma.; if the
bias is increased to 6 volts the plate current decreases to 2.4 ma.
These points on the static characteristic curve are labeled as A,
B, and C.
These points are called the operating points. Thus when the
C bias is changed the operating point slides up and down on this
steep H,-I, curve. If this C bias is changed in some regular
fashion, say in the form of a sine wave between 4 and 6 volts peak,
the plate current must change accordingly, that is, it will increase
and decrease between 3.6 ma. and 2.4 ma., as the curve shows
The average bias is 5 volts; the average current is 3.0 ma. The
maximum bias is 6 volts, the minimum is 4; the maximum current
is 3.6 ma., the minimum is 2.4.
202
THE TUBE AS AN AMPLIFIER 203
The bias may be changed in this manner by setting up an
a.-c. voltage across a resistor as in Fig. 139. Suppose, as an example
the voltage, E, across this resistor is a sine wave of a maximum
value of one volt. When this sine wave of voltage makes the grid
end of the resistor negative by one volt the actual voltage on the
5
E. or Bias Voltage E
Fig. 138.—Use of straight part of characteristic. The wave form of output
current will be exactly similar to the wave form of the input voltage if the
characteristic is straight.
grid will be 5 volts (Z.) plus 1 volt (Z) or 6 volts in all. The plate
current will be, at that instant, 2.4 milliamperes. When the a.-c.
voltage reverses and makes the bottom or filament side of the
resistor negative, the bias on the grid is 5 volts (#,) minus 1 volt
(EB) or 4 volts in all; the plate current at this instant will be 3.6
204 THE TUBE AS AN AMPLIFIER
ma. At all other instants the plate current will be different
depending upon the instantaneous value of the a.-c. input voltage.
The plate current changes in unison with the grid voltage, and
if this grid voltage varies in a sine wave there will be a sine wave
of current in the plate circuit.
We can picture what happens by the sine waves in Fig. 138.
With no input the grid voltage is that of the C battery, the plate
current is that corresponding to this value of C bias and a plate
voltage, E,. Now let us assume that an a.-c. voltage requiring
one second for a complete cycle is applied to this grid. Starting
from zero it increases in a
positive direction making
the grid more positive (less
negative) with respect to
the filament. The plate
current begins to rise ac-
cordingly. At the end of a
quarter-second (90°) the
Fig. 139.—When £; is applied, J, consists voltage is a maximum in
of an a.c. current as well as steady plate the positive direction and
current. the plate current is corres-
pondingly amaximum. Now
the grid voltage begins to decrease (become more negative), the plate
current decreases, and at the end of a half-second (180°) the voltage
and current arrive at their original no-input values. Now the grid
voltage (H, = EH. + EF) continues to decrease until at the three-
quarter-second phase the voltage is a minimum and so is the plate
current. Thereafter the voltage and current increase again and at
the end of a full second are equal in value to the starting or no-
signal values. Thereupon the cycle is repeated.
Notice that the two curves have the same form; that the steady
bias line of 5 volts is in the exact center of the a.-c. input voltage
curve; that the average plate current is the current corresponding
to 5 volts negative C bias. The fact that the plate current curve
seems to have less amplitude than the grid voltage variation is not
important—this is a matter of the scale to which the two are
drawn. One is in volts, the other is in milliamperes. This a.-c.
RESISTANCE OUTPUT LOAD 205
plate current may be thought of as a variation in the d.-c. plate
current at a rate corresponding to the frequency of the input
voltage, or as a true a.-c. current which flows in the plate circuit
in addition to the d.-c. current.
A meter in the plate circuit of the tube would read only the
average current unless its needle were capable of following the
changes in current. If the input a.-c. voltage were 1000 cycles,
for example, the needle would not be able to follow such rapid
variations and would register only the average value of plate
current, that is, the plate current corresponding to 5 volts bias or
3.0 milliamperes.
This is essentially the theory of the action of the tube as an
amplifier. An input a.-c. voltage is applied to the grid. The
variations in grid voltage about some average value equal to the
C bias produce corresponding variations in plate current. These
variations in current are caused to flow through some sort of output
load impedance, and across this impedance they set up voltage
variations. If the bias (#,) and plate voltage H, are properly
chosen so that the variations in H, (dH,) take place on a straight
part of the #,-I, curve the form of the a.-c. current wave in the
plate circuit will be exactly similar to the a.-c. voltage wave on
the grid. If the proper conditions of E., EH, and load impedance
are fulfilled, the voltage appearing across this impedance will be
not only an exact replica of the a.-c. grid input voltages, but will
be an amplified replica of them.
The question naturally arises, how much amplification can be
obtained? What are the conditions for such maximum amplifi-
cation?
175. Resistance output load.—If, in the plate circuit of the
tube we put a resistance, Ro, as in Fig. 140, we may adjust condi-
tions so that distortionless amplification results. These conditions,
briefly, are: (a) the C bias and magnitude of the input a.-c. volt-
age must be such that only the straight part of the characteristic
is used; and (b) the load resistance, R., must be large compared
to the plate resistance of the tube.
Let us look at the circuit in Fig. 140 rather criticially. It is
the fundamental amplifier circuit. The plate voltage, H,, as
206 THE TUBE AS AN AMPLIFIER
measured by a voltmeter connected from plate to the negative
filament lead is no longer the voltage across the B battery. It is
less than this value by the voltage drop in the resistor Ro, i.e.,
the IR drop caused by the plate current. The voltage actually on
the plate then is
E, = Ey — Ip X Ro (1)
If, for example, the B battery voltage = 180 volts, Ro =
100,000 ohms, J, = 0.5 milliampere, J, X Ro = 0.0005 x 100,000
or 50 volts, and Z, = 180 — 50 = 130 volts.
Now it can be seen that any variation in J, causes a variation
in the JR drop across Ro, and hence the plate voltage HL, must
change according to equation
(1). For this reason any varia-
tion in the grid bias will cause
a variation, not only in the
plate current of the tube but
a variation in the plate voltage
as well. We cannot use the
static characteristic to deter-
mine what the output voltage
Fig. 140.—Ratio of EZ, to E; isthe volt- looks like, because the plate
age amplication of circuit. voltage is no longer constant,
but changes in instantaneous
value with each instantaneous value of the grid voltage. The
operating point, then, does not slide up and down the charac-
teristic curve as plotted in Fig. 138 but along a new kind of
curve known as a dynamic characteristic curve.
176. Dynamic characteristic curves.—A series of dynamic
curves is shown in Fig. 141. They were taken by placing resist-
ances in series with the plate battery and a 171 power tube main-
taining the voltage on the plate equal to 180 when the C bias was
38.5. It will be noted that they are much flatter and longer
than the static curves. This means that plate current variations
are much smaller in magnitude under the same grid voltage varia-
tions. The mutual conductance of the circuit is no longer as high as
the value for the tube alone—but the slope of the curve tells us the
PHASE OF E,, E,, AND I, 207
a.-c. current which will flow through the resistor when a given a.-c.
voltage is applied to the grid, and these curves are therefore more
useful than the static curves.
Experiment 1-10. Connect as in Fig. 140 a tube of the 201-A type, a
plate battery, and a resistance. Short-circuit terminals for E;. Measure
and plot the current in the plate circuit as the grid voltage is varied. Then
change the resistor and repeat. Use values of 8,000, 16,000, 24,000 and
48,000 ohms.
177. Phase of E,, E,, and I,.—When the grid of the tube is
made negative the plate current decreases; when the grid is made
Radio Broadcast
Laboratory
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“*K
SN
~
8000 Ohms
Plate Current
0 10 20 30 40 50 60 70 80 90
Grid Voltage
Fig. 141.—Dynamic characteristic curves.
less negative the plate current increases. When the plate current
increases, however, the voltage drop, /R., across the resistor, Ro,
increases and for this reason the voltage H, actually on the plate
decreases. That is, the plate current increases when the grid
voltage increases (becomes less negative), but on the contrary the
plate voltage decreases when the grid voltage increases. Thus we
may say that the plate current variations are in phase with the
grid voltage variations whereas the plate voltage variations are
208 THE TUBE AS AN AMPLIFIER
out of phase with the grid circuit variations. These phase relations
are shown in Fig. 142.
178. Magnitude of the amplified voltage.—There are several
variable factors in a one-stage resistance-coupled amplifier, as
shown in Fig. 140. The grid bias, E., the battery voltage, Ho, the
Increase
™m
7°
E, With no
input ég
Increase
Decrease
I
Tp With no
input eg
Magnitude
Increase
Decrease
m
s
Decrease
Fig. 142.—Phase relations between a.-c. grid and plate voltage and plate
current.
load resistance, Ro, all may be changed as desired. With a fixed
value of H, and EH», the plate resistance of the tube, R,, is fixed.
If we go into the laboratory and measure the voltage across a load
resistance with a given value of a.-c. voltage on the grid we shall
determine the voltage amplification of the stage. Then if we
change the load resistance and measure the output voltage across
R we shall get a curve which shows that the amplification increases
EQUIVALENT TUBE CIRCUIT 209
as the toad resistance increases, but finally comes very near a
certain fixed value—which is numerically equal to the amplifica-
tion factor of the tube—and that increasing the load resistance
beyond this point has little effect on the amplification.
Because each change in Rp will change the plate current, we
must adjust the plate battery each time so that the voltage actually
on the plate—which is EZ, minus the voltage drop along the load
resistance—is the same.
The amplification that will be realized in a laboratory experi-
ment of this kind may be calculated from this formula
uRo
Tees = @ (voltage amplification) (2)
and the r.m.s. a.-c. plate current from
Meg
oot ees £5 3
aa, ee (3)
and the r.m.s. a.-c. voltage across the load from
be, Ro
= Ge, = &, 4)
Rot Rp : \
where e, = r.m.s. grid voltage.
179. Equivalent tube circuit.—These same voltages and cur-
rents can be obtained from the circuit in Fig. 143 by simple Ohm’s
law calculations. For this reason it is
standard practice to substitute for the
tube its equivalent consisting of a voltage
uwE, in series with two resistances, one
equal to R,, the tube resistance, and
the other, Ro, equal to load resist-
ance.
- The maximum voltage amplification Fig 148 Bquivelon ae
takes place when the external resistance, Ro, Gait of amplifier tube!
is infinite. Practically, however, 75 per
cent of the maximum amplification is obtained when A, is 3 times
210 THE TUBE AS AN AMPLIFIER
as great as R,. For example, if R, is 10,000 ohms, a coupling
resistance of 30,000 ohms will realize 75 per cent of the » of the
tube. The maximum possible value is the amplification factor of
the tube. Thus if a tube especially made for resistance-coupled
amplifiers having an amplification factor of 30 is used, the maxi-
mum possible amplification will be 30, but under actual operating
conditions the voltage amplification, K, of the complete circuit is
about 20.
180. Power output.—The a.-c. power in the load resistance is
calculated as follows,
Pieter:
é Mey
by
Retake
(uey)* Ro
=i pio ee (5)
(Ro + Ry)
This is a maximum when the two resistances are equal, that is,
when
R, = Rao
Then the power (Paes Hees
DERE 4R, (6)
where e, = r.m.s. input voltage;
and
2H 2
2) Se etl =
SR, (7)
where H, = peak or maximum input voltage.
This power which is fed into the load resistance must come
from the plate battery, because the tube itself generates no power
—it acts merely as a transformer or valve which takes small
voltages and currents on its input and turns out larger voltages
and currents to its output circuit. The power in the input circuit
must come from the circuit to which the tube is attached. The
power in the output must come from the batteries. The tube
therefore releases power from the batteries in a form which is an
POWER AMPLIFICATION 211
exact replica of the power utilized in the input circuit to which the
tube is attached.
181. Power amplification.—Ordinarily the grid of a tube is
biased so highly negatively that practically no current flows in the
grid circuit. The tube input circuit itself then draws little or no
power. If, however, the a.-c. voltages are fed into the tube through
a resistance, some power is expended in this input resistance. If
the values of resistance and current are known the power may be
calculated. The power amplification will be the output power
divided by the input power. The fact that a tube is a power
multiplier distinguishes it from a transformer which is merely a
power transmitter, taking power at one current and voltage and
passing it on at another current and voltage. The transmitted
power is never more than or even equal to the input power. It is
always less; unlike a tube, the transformer cannot amplify power
or release it from local batteries. The power output is propor-
tional to the square of the voltage on the grid. Thus doubling the
input a.-c. voltage quadruples the output a.-c. power.
Example 1-10. Four milliamperes of current at 1000 cycles are fed through
a 10,000-ohm resistance in series with the grid of a tube and its C battery.
In the output is a 2000-ohm resistor. The amplification factor of the tube
is 3, its internal plate resistance is 2000 ohms. What is the power output,
what is the voltage across the output, what is the voltage and power amplifica-
tion, and what power is lost on the internal resistance of the tube?
Let the input voltage EZ; = [;R; = .004 X 10,000 = 40 volts r.m.s.
uh
ip Ss
voltage amplification G =
3 & 2000
~ 4000
= 1.5 times.
The output voltage HE, = H; X G= 40 X 1.5 = 60 volts r.m.s.
P tput P Ly
ower outpu =
x 4Ry
9 < 40?
= —— = 1.80 watts
4 X 2000
212 THE TUBE AS AN AMPLIFIER
Power input P; 1;2R; = (.004)? & 10,000
= 16x 10-8 / 10,000
= .16 watt
Power amplification ioe — 1.3) lor—slle2o) times:
Since I, (a.c.) flows through both R, and Ry and since Rp = Rp a.-c. power
lost in R, = power lost in Ro = 1.8 watts.
Total power taken from batteries = 3.6 watts.
Effici useful power _ 1.8 Bey Norcent
ee eternal power 3.6 - i
Problem 1-10. Assume a 201-A tube with a load resistance of 12,000
ohms and an a.-c. grid voltage of 3 volts (peak). The u of the tube is 8, its
R, is 12,000 ohms. What is the maximum value of the a.-c. plate current?
(Use formula 3.) What is the r.m.s. value? What a.-c. power is developed
in the load resistance? What voltage appears there?
Problem 2-10. What is the voltage amplification in the above circuit?
(Use formula 2.) How much would it be increased if the load resistance were
increased. to 36,000 ohms? How much would this change the power output?
Problem 3-10. ‘The plate resistance of a certain “high w”’ tube is 60,000
ohms and its amplification factor is 20. What plate resistor value in ohms
must be used to realize a voltage amplification of 15?
Problem 4-10. A 112-type tube has an internal plate resistance of 5000
ohms, its amplification factor is 8; it is worked into a load of 5000 ohms. Plot
the output power in the load resistance as the input a.-c. voltage is increased
from 1 to 8 volts (peak). (Use formula 7.)
Problem 5-10. What is the voltage amplication in Problem 4.
Problem 6-10. The power output of a tube when worked into a load
whose resistance is equal to the tube resistance (when the input grid voltages
2
are maximum volts) may be written as a x H,*. If we divide this expres-
sion by H,? we shall have a figure which gives us the power output in watts
per (volt input)?. Make a table of such values for all the tubes used at the
present time, getting the tube data from the tube chart.
Problem 7-10. The normal C bias for a 112-type tube is 9 volts. What
is the largest r.m.s. voltage that may be applied to its grid before the grid goes
positive?
182. Amplifier overloading.—The conditions for undistortea
amplification are: (a) the C bias and magnitude of the a.-c. input
voltage must be such that only the straight part of the tube’s
AMPLIFIER OVERLOADING 213
characteristic is used, and (b) the load resistance must be large with
respect to the internal resistance of the tube, R,. Let us examine
these conditions. Suppose first that the C bias is too great, as in
Fig. 144, so that the operating point goes into a curved part of the
characteristic. The wave of plate current is no longer similar to
that of the grid voltage, and its average value is no longer equal
A.C. Plate Current
Increase in Plate Current
Normal D.C. Plate Current
Fig. 144.—Effect of using curved part of characteristic. Note dissimilarity
between output and input.
to the zero a.-c. input condition as is true when the bias is such
that input voltages carry the grid operating point only over a
straight part of the curve. The negative parts of the a.-c. waves
are partly cut off. Distortion results because the negative and
positive halves of the cycle are not amplified alike. A meter in the
plate circuit would show an increase in current when signals were
put on the grid—an infallible sign of overloading distortion. If
the C bias is great enough and input signals are strong enough
214 THE TUBE AS AN AMPLIFIER
the grid may be forced so far negative that the plate current may
be reduced to zero on the negative halves of the wave. This would
produce even worse distortion.
Overloading, then, is a technical name for operating a tube
under a wrong C bias, or with too strong input signals.
If the C bias is too small, input signals force the grid positive
at times and distortion again occurs for a reason explained in
Section 185. In this case the plate current as read by a d.-c.
meter would decrease when the input voltages are applied to the
grid. : |
Suppose, however, the tube is biased properly, say at the
center of the region between the lower bend of the H,-I, curve and
the point which corresponds to zero grid voltage. Distortion does
not occur unless too great a.-c. voltages are put on the grid—
voltages sufficient to drive the operating point down on the curve,
or up on the positive part of the curve.
There are two possible remedies for such kinds of distortion.
One (a) is to reduce the input voltage until the operating point
moves over only a straight portion of the curve; another (6) is to
increase the B and C voltages until a longer straight portion is
available. As shown in Fig. 118, increasing the plate voltage moves
the #,-I, curve to the left and increases its straight part. The C
bias would be increased accordingly.
If the tube is properly biased, the input a.-c. voltage must be
such that its peak value does not exceed the value of the C bias
voltage. Thus if a tube has a bias of 40.5 volts, the input value
of the a.-c. voltage must not exceed this value, and thus its r.m.s.
value must be not over 40.5 + 1.4, or about 28.5 volts. A voltage
input greater than this will force the grid positive with consequent
distortion. As seen in Section 186, the peak voltage input must be
even less than this value to prevent the operating point going
down into this curved region.
183. Distortion due to curved characteristic—When a load is
in the plate circuit of a tube its characteristic becomes flatter and
its straight part longer, but at the lower part of the curve there is
still a bend and throughout the E,-I, graph there is considerable
curvature unless the load resistance is high. Large input voltages,
DISTORTION DUE TO CURVED CHARACTERISTIC 215
therefore, may cause distortion because the operating point
traverses curved parts of the characteristic.
Although the maximum power output is obtained from a tube
when the load resistance is equal to the internal tube resistance,
the maximum undistorted power output is attained when the load
resistance FR, is twice as great as the tube resistance, Rp. Some
power is sacrificed under these conditions but, as the curve in
Fig. 145 shows, the loss is not great until the output load resistance
is several times the tube resistance. If the load resistance is lower
14
=
No
=
o
Undistorted Power Output (Watts)
.06
04 —— 4
CX 112-A
ie E,=135 Volts |
E.=-9.0 «
Rp=4500 Ohms
4000 8000 12000 16000 20000 24000
Load Resistance (Ohms)
Fig. 145.—Output power as a function of load resistance.
than the tube resistance distortion due to curvature of the charac-
eristic is bound to occur. Even with a load equal to twice the
tube resistance there is considerable curvature at the bottom of
the characteristic and so input voltages high enough to drive the
plate current too low must not be applied. See Section 186. The
power output when A. = 2 #, is
2 prey”
SS Seuss 8
P a R, (8)
Distortion frequently occurs at low audio frequencies. There
are two reasons for this. In the first place the excursions of the
operating point may be somewhat greater on these frequencies
216 THE TUBE AS AN AMPLIFIER
because of the greater power in them than in high tones. In the
second place, the loud speaker may have a resistance which varies
with frequency, becoming low at low frequencies, probably lower
than the tube resistance. Thus distortion due to curvature as
well as loading distortion may occur on low notes.
The solution is to use a power tube which will easily handle the
greatest input grid voltages that will be encountered, and secondly,
to use a power tube with a low internal resistance so that the load
resistance will, at the lowest frequency, be larger than the tube
resistance.
184. Permissible grid swing.—The expression “ grid swing ”’
is often used, especially in England, to indicate the extent to which
the voltage of the grid varies under the incoming signals. To state
that the grid swing on a certain tube is 10 volts means that the
voltage of the grid varies between 5 above and 5 below some fixed
value, a total swing of 10 volts. In this country we should say the
maximum input voltage was 5. Under these conditions the
C bias should be 5 volts at least, preferably more. The maximum
permissible grid swing is the range of voltage on the grid which
will not cause distortion either because of the grid going positive
or because of the operating point traversing the lower bend.
Whether one says the grid swing that may be applied to a 171
power tube is 80 volts or whether he says that the peak voltage
that may be applied is 40 volts is immaterial. They mean the
same thing. ‘The grid swing cannot be determined by looking
at the static characteristic curve. It can be determined from the
dynamic curve or as indicated in Section 186.
185. Distortion due to positive grid—Why does distortion
occur when the grid of an amplifier tube is permitted to swing
positive on loud signals? If we plot the plate current against grid
voltage, as the C bias in Fig. 139 is changed, we find that, when
the grid goes positive, the plate current curve no longer is straight;
it slumps off and soon becomes almost horizontal. Distortion
will occur at the instant the input voltage takes the operating
point up on the upper bend.
The reason for this bend is as follows. When the grid becomes
positive with respect to the negative side of the filament, it begins
‘
AMOUNT OF DISTORTION CAUSED BY OVERLOADING 217
to attract electrons to it and these electrons constitute a flow of
current. Current flows in the grid circuit, and must go through
the input resistance R;. There is, then, an JR drop in the grid
circuit, so that the voltage actually on the grid (EZ,) is not the
applied grid voltage H but this value minus the drop in the input
resistance—just as the plate voltage is not the voltage of the B
battery but this voltage minus the drop in the output load resist-
ance. The greater the input voltage the more the grid goes
positive, the greater the voltage drop in this resistance and the
smaller the proportion of voltage that is actually on the grid.
The grid must never be permitted to go positive in ordinary
circuits. Some circuits have been developed in which the grid is
not only permitted to go positive but is forced to do so with a
consequent greater power output. Such circuits are not, as yet,
in general use.
186. Amount of distortion caused by overloading.—What is
the result of operating amplifier tubes over a curved part of the
characteristic? The result is an output wave form quite different
from the input wave. Distortion is taking place whenever the
magnified output voltage is not exactly, in all respects, like the
input voltage. When the d.-c. plate current of an amplifier tube
changes under the action of an input a.-c. grid voltage, distortion
is taking place; the form of the wave in the output circuit does
not look like the form of the input wave. This output wave may
be very complex.
When a tube distorts, it adds certain frequencies to the output
circuit which were not present in the input. All of these frequencies
added together at any instant produce a wave form which looks
unlike the original wave form.
How can we tell the amount of distortion to be expected with
given tube and circuit constants? It is not difficult to determine
what percentage of distortion will occur. Here is another place
where the characteristic curves come in handy; this time the
E,-I, curves are used.
Let us look at Fig. 146 which gives the characteristic of a
Cunningham CX~371 type of tube. These data are taken from
the Cunningham Tube Data Book of 1927. The vertical 180-volt
218 THE TUBE AS AN AMPLIFIER
line is the working line for this tube. The intersection of this line
with the 40-volt grid bias line gives the plate current, 19 milli-
amperes. If an a.-c. voltage is placed on the grid—in addition to
the 40-volt d.-c. bias voltage, of course—say of a maximum or
peak value of 10 volts, the actual grid voltage will vary 10 volts
up and 10 volts down from the 40-volt bias; that is, at one instant
Plate Current
i ag 140 ere ami 200 220 240
Plate Voltage
Fic. 146.—Static characteristic of power tube. Dashed lines represent
characteristic with a load in the plate circuit.
it will be —30 volts and at another —50 volts. Looking at the
appropriate curves in Fig. 146 we see that when the grid voltage
E, is —30 the plate current is 38 ma. and when the voltage is —50
the plate current is 6 ma. These represent changes of 19 ma.
(38 — 19) and 13 ma. (19 — 6) respectively. Although the input
voltage is symmetrical about the 40-volt bias voltage, the plate
AMOUNT OF DISTORTION CAUSED BY OVERLOADING 219
current is obviously not symmetrical—there is a greater change in
current (19 ma.) in one direction than there is in the other (13 ma.).
The output curve would not be a magnified replica of the input
curve. Harmonics would appear in the output.
Now when a load resistance is placed in the plate circuit, the
straight part of the curve becomes longer and flatter as shown
in Fig. 141, and as more and more resistance is added the actual
El=0 10 —20 30 Ly
Calculation of Slope of Load Line
55 CX — 9371
YA=180—130=50 volts —650
50 XY= 31 —18=13 ae:
245 XV/yA = 13/59 as lf, 0 =3900
§
#40 6
2 = 60
2 3m
z ‘ r Z
30 oe
! W
25 Kn
|
20 $f A 4
15 y
10 4 :
8
5 es |
I|min, B
0
80 100 120 140 160 180 200 220 240 260
Plate Volts
Fic. 147.—Method of plotting “load line”’ of a power tube. From such a
curve all of an amplifier’s performance can be determined.
working lines rotate about the intersection of the 180-volt plate
voltage line and the grid bias line which for this curve was 38.5
volts. The plate current changes are no longer as great but
they occur symmetrically about the normal plate current value
which is governed by the plate voltage and the bias voltage. Thus
for a given value of load resistance such that the dotted lines in
Fig. 146 represent the plate current as controlled by plate and
220 THE TUBE AS AN AMPLIFIER
grid voltages, the same input voltage 10 (peak) causes the plate
current to increase from 19 to 24 ma. in one direction and to
decrease from 19 to 14 ma. in the other—a symmetrical varia-
tion.
It is a simpler process to determine the plate current variations
with a given plate load resistance by the method of plotting the
“load line.”’ To do this a family of #,—I, curves is necessary as
in Fig. 147. The load line gives the plate current at any given
set of EH, and EH, values and with a given resistance load. It is
determined as follows: it must pass through the intersection of the
E, = 180 and FE, = — 40.5 lines, or A in the figure, because these
are the recommended values for this tube, a CX-371. Through
this point A draw a line parallel with the plate voltage axis.
Make it any convenient length. Then at the end of this line
erect a perpendicular line of such length that YX /AY= 1000/R,
where F, is the plate load resistance, in this case 3900 ohms.
Through X draw a line through A and extend to the H, = 0
line and the J, = 0 line as in the Figure.
The reasons for this bit of apparent sleight-of-hand are as
follows. The slope of this load line must be the reciprocal of the
load resistance, i.e., if the load resistance, R,, is 3900 ohms, the
slope of this line must be 1/3900 ohms. ‘The slope of the line will
be equal to the vertical side of a triangle divided by the horizontal
side of this triangle which has for the third side either the load
line, or a shorter line parallel to the load line.
Thus, slope of load line:
vertical 1 I(amps.) _ I X 1000 ma.
horizontal R, & (volts) — E volts
The value of plate current corresponding to any set of H, and
FE, values with this particular load resistance may be found from
this line. Thus with #, = 0 and E, = 92 volts, J, = 39 ma., with
E, = —60 and E, = 218, J, = 9 ma. Thus if the bias voltage
is 40.5 and the peak voltage applied to the grid is 10, the actual
grid voltage varies from 40.5 — 10 or 30.5 to 40.5 + 10 or 50.5 and
the corresponding plate currents will be 24 and 13 approximately,
or approximately 5.5 ma. up and down from the 18.5 ma. value.
HARMONIC DISTORTION CALCULATION 221
Now the grid must not go positive, and the plate current must
not become so low on strong signals that the curved part of the
characteristic is used. Thus 1.0 ma. is about the lowest the plate
current should go, or the value when ZH, = —80. From —40.5
to —80 is 39.5 volts which is the peak a.c. that should be put on
the grid. This input will cause the plate current to vary between
39 ma. and 1 ma.
187. Power output calculation—The power output may be
calculated from the above data by means of the following formula:
P = 1/8 (# max. — E min.) X (I, max. + I, min.)
1/8 (250 — 92) x (.039 — .001)
= .75 watt = 750 milliwatts of output power.
188. Harmonic distortion calculation.—The amount of distor-
tion present in such an amplifier working under such conditions
may be found from the following equation in which J, = 18.5 ma.
which is equal to the plate current when 180 volts are on the plate
and the C bias is 40.5 volts.
(1, max. + J, min.) — I
iH
Distortion = Gea EAE
& 4 (.0389 + .001) — .0185
7 .039 — .001
eu .02 — .0185 LA
.038
= 3.9 per cent.
This distortion is the amount of second harmonic current in
the output. Thus if the input a.-c. grid voltage has a frequency
of 1000 cycles and if the output current is 10 milliamperes (fun-
damental) there will be in the plate circuit 3.9 per cent of this
current or about 0.4 milliampere in the form of a 2000-cycle wave.
It has been determined that a 5 per cent distortion is not objec-
tionable to the ear—but this involves a matter of opinion and other
factors. If and when better loud speakers are available it is
possible that the average car will detect less distortion than this.
222 THE TUBE AS AN AMPLIFIER
189. Power diagrams.—To review the tube with a resistance
load as a power amplifier, and to gather a few additional facts, let
us draw the E,-I, curves as shown in Fig. 148, which is a purely
theoretical case—the curves and values of current and voltage
represent no particular tube now available; they were chosen at
random. Let us assume that the voltage actually on the plate is
160 volts when the steady plate current is 20 ma. That is, H, =
160, and J, = 20 ma. Let us assume a load of 5000 ohms, find
Eg=Ep, +Ep
Fig. 148.—Power diagram. The area of shaded triangle represents a.-c.
power used in load.
the slope, and draw the load line through the intersection of
the I, = 20-ma. line with the #, = 160-volt line, and find that
the bias on the grid to maintain such a plate current is 20 volts.
Let us assume that a sine wave is put on the grid whose maximum
value is 10 volts, so that the actual grid voltage varies between
— 10 and — 30 volts. The corresponding J, variations are from
30 to 10 milliamperes.
We note that the load line crosses the battery voltage line at
about 260 volts. This, then, is the battery voltage necessary to
POWER DIAGRAMS 223
insure that the plate of the tube gets its 160 volts. In other words
there is a drop of 100 volts (5000 X .02) in the resistance of the
load, R. Under normal conditions of no a.-c. input to the grid,
the voltage across the tube is 160 volts, across the load is 100
volts, and the plate current is 20 ma.
Now the d.-c. power lost in the load is the product of the volt-
age across the resistance R and the current through it; and the
power lost on the plate of the tube is the product of the voltage
across the tube and the current flowing. Thus,
Bee Lee DE CCD
Bae OD COD
Power lost in load
I|
Power lost in tube
and the total power supplied by the battery must be the sum of
these powers, that is
Power from battery = Es X I, = OE X CD
Now the area of the rectangle CDEF = DE X CD = power
in load, and the area of the rectangle ODCG = OD X CD = power
lost in tube, and the area of the rectangle OFFG = OE X CD =
total power from battery. ,
When an a.-c. input is applied to the tube so that the grid is
operated about its mean or average value of — 20 volts, the
maximum a.-c. current through the load is given by the line HC
and the maximum a.-c. voltage across the load is given by the line
HB. Let us call these values of current and voltage
érz = maximum a.-c. voltage across load = HB
ip = maximum a.-c. current through load = HC
Since a.-c. power in a resistance circuit is the product of the
r.m.s. current and the voltage, to obtain the a.-c. power in the
load we must first get the r.m.s. values of the above values and
then multiply them
é€rY.M.s. = @k Max. x er
2,
ty Y.M.8. = tp MAX. X we
2
224 THE TUBE AS AN AMPLIFIER
whence a.-c. power in load = er 7, (r.m.s.)
1 i]
= ae Sa
er X ty
2
ESO
5 = area of triangle HCB.
This power is dissipated in the load resistance in addition to
the d.-c. power lost there due the voltage drop across it and the
current through it. Since the average current drawn from the
battery has not changed, the power taken from the battery has
not changed—and yet the load has an additional amount of power
used up in it. Where does this power come from? Clearly it must
come from the power used up on the plate of the tube. When an
a.-c. voltage 7s placed on the grid, then a.-c. power is developed in the
load, less power is wasted on the tube plate, and the tube will actually
run cooler when it 1s delivering power to the load than when standing
idle, that 1s, with no a.-c. input grid voltage.
Let us see what these values of power are. We can take them
directly from the graph in Fig. 148.
d.-c. power lost in load (no a.-c. grid voltage)
= DE X CD = (260 — 160) X (20 ma.)
= 100 X .02 = 2 watts
d.-c. power lost in tube (no a.-c. grid voltage)
=OD < CD = (160). <' (20 ma.)
= 160 X .02 = 3.2 watts
d.-c. power taken from battery
= OE X CD = 260 X .02 = 5.2 watts
max. a.-c. voltage across a load (a.-c. grid voltage = 10)
=HB = 210 — 160 = 50 volts
POWER DIAGRAMS 225
max. a.-c. current through load (a.-c. grid voltage = 10)
= HC = 10 ma. = .01 ampere
a.-c. power in load (a.-c. grid voltage = 10)
we BOC HC 7.59650 X7.01) = 425 watt
= 250 milliwatts, which is subtracted from d.-c. power in tube.
In Section 180 the formula for the DON output of a tube was
stated to be
pe?R
(R, + R)?
power output =
and in order to check the above calculations we must find the
tube constants from the curves in Fig. 148. The plate resistance of
the tube is simply the reciprocal of the slope of the E,—I, line, and
noting that a change of 160 — 100 volts on the #, =— 10 line
‘produces a change in plate current of from 42 to 25 ma. we
calculate
160 — 100 60
BA Se ete 50) oli
Ry = “O49 — 025 017 ae a
and from the two curves, #, = — 10 and H, = — 20 we ascertain
that a change of 10 volts on the grid produces a change in plate
current of from 42 to 20 ma. when the plate voltage is 160, we
calculate
Tp = ee = 2200 micromhos
10
whence
Rp X Gm = 3500 X 2200 X 10%
= 7.7
=
I
whence power output
_ (7.7 X 7.07)? X 5000
(5000 + 3500)?
.206 watt = 206 milliwatts.
l|
226 THE TUBE AS AN AMPLIFIER
This value agrees closely enough with our data secured from
the characteristic curves in Fig. 148.
Thus from a collection of the H,—I, curves of any tube we may
obtain all the necessary data upon which to build an amplifier and
calculate its power output, the losses in the various tubes, the
percentage distortion, etc.
190. The pentode.—A triple-grid tube which has come into
general use in Europe and which is coming to some prominence
A, Es,
s Py) =Power Output
33
u
oe
°
Aa
i afP Power
| | B = Egi Sensitivity
Ip =15 ma. Fa
sg 3. .
10,000 20,000 80,000 40,000
R ,=Load Impedance —— Ohms
Fic. 149.—Power output and harmonic content of output of pentode tube.
in this country is the pentode, a power output tube. One grid is
permanently attached, within the tube, to the filament, another
is connected to the plate voltage, and the third is the signal grid.
The cathode grid (attached to the filament) forms a wall through
which electrons emitted from the plate due to impact of normal
electrons from filament, cannot break through to reach the other
elements.
The advantage of the tube is its superior sensitivity compared
to a three-element power tube. With comparatively small input
THE PENTODE 227
voltages considerable power output can be secured. Its disad-
vantage is that considerable third harmonic distortion exists in its
plate circuit. The tube is a high-mu, high plate resistance tube,
and should work into a load resistance considerably less in value
than the resistance of the tube.
Wherever a tube giving considerable output with small signal
input is desired, e.g., automobile radios, police receivers, airplane
sets, etc., the pentode fits in. It is a space-saving, amplification-
saving tube. It is probable that many hundreds of them will find
their way into radio receivers for special purposes and perhaps be
generally used in home receivers when the technical problems
connected with their use have been completely studied and solved.
The tube is specially valuable when detectors overload. Using
a pentode instead of a three-element power tube, the detector
need not put out as much voltage, and hence its input may have
to handle smaller voltages.
Experiment 2-10. Remembering that the effective plate voltage may be
calculated from
H = EH, + wy
and that E, is negative use the data below to plot a family of Ey-I, curves.
A good way to do this is to fill in the table below with the essential data and
then to plot it. After two or three curves are plotted, the others may be
drawn without calculating the values of current because all curves are parallel.
Calculate the mutual conductance, the amplification factor, the plate resist-
ance. Calculate the slope of the load line (AZ in Fig. 148); draw it in, and
figuring 5 ma. as the minimum permissible value of plate current, calculate the
maximum and minimum voltage across and current through the load, the
power in it, the percentage distortion. Assume the efficiency is the ratio
between a.-c. power in load and d.-c. power taken from the battery. Calculate
the efficiency for several values of input a.-c. voltage.
Ep E,=-0 £,=-10 H,=—-20 F,=—30
150 20 Ue hear e || ccmsori otc
200 30.2 EGY” @ Meech Moat
250 42 29). Omit ) W) Meeye ear:
300 is 5) Wee aactndie mL” f berate ca a) We gereron
350 COMME A A certs | Mais teres
ee, me CHAPTER XI
— AUDIO AMPLIFIERS
“ So Far we have not spoken of the frequency or band of fre-
quencies at which the vacuum tube and its associated apparatus
will amplify. The theory up to the present point deals with
amplifiers in general. It is necessary now to consider the types of
amplifiers, and their differences.
191. Need of an audio amplifier—Practically all radio re-
ceivers in use at the present time consist of three parts: a radio-
frequency amplifier, a detector, and an audio amplifier. Because
of the simplicity of audio amplifiers, and because of their greater
application, we shall consider them first.
Let us think of the broadcasting studio in which originate the
signals which we must amplify. A microphone stands in front of
a musician or a speaker. It has two electrodes which for the sake
of our discussion may be made of carbon. Between them is a
little box of carbon granules. One of the electrodes has a metal
diaphragm attached to it which is so located that the air vibrations
we call sound impinge on it. When sound is directed into the
microphone it moves the diaphragm which in turn moves one of
the electrodes which squeezes the carbon granules and thereby
changes their electrical resistance. The steady battery current
which normally flows through the microphone from one carbon
electrode through the carbon granules to the other electrode is
changed in accordance to the frequencies of the music or speech
directed into it.
This steady current of the microphone is said to be “ modu-
lated ” or changed in accordance with the voice frequencies; and
after the steady current is filtered out, the changes are built up in
strength by amplifiers which will transmit the range of frequencies
228
NEED OF AN AUDIO AMPLIFIER 229
that will be encountered in practice. Here then is the first need
for amplifiers. They must make up the loss resulting from trans-
ferring the energy in the moving air particles we call sound into
the energy of the changing electric current. This loss in energy
that must be made up is considerable.
The frequencies normally transmitted over telephone lines
range from 250 to about 2500 cycles. These are complex audio-
frequency a.-c. currents. Most of the energy of the voice occurs
in the frequencies below 1000 cycles, most of the intelligibility
above that frequency; therefore if a filter is put in the telephone
line which admits only frequencies below 1000 cycles we would
hear a sound but it would be unintelligible. On the other hand,
if the filter cut out all the low tones and transmitted only those
above 1000 cycles, we could understand what the speaker was say-
ing, but the sound would not carry, it would be weak.
For best intelligibility and carrying power—naturalness, we
say—all the frequencies from about 120 to 2500 are necessary for
transmission of speech.
Music, however, is more complex and for realism and natural-
ness a much greater frequency range must be transmitted, not only
by the microphone but also by the amplifiers, the radio broadcast-
ing station, the ether, the radio-frequency amplifier or the receiver,
and finally the audio amplifier and loud speaker in one’s home.
Music requires the transmission of all frequencies between 100 and
5000 cycles per second, and many critics—especially in HEurope—
desire that a still greater range shall be transmitted. At the present
time, in the United States, the best broadcasting stations and con-
necting circuits transmit from slightly below 100 to about 5000
cycles per second. After a certain amount of amplification these
tones are mixed with a radio-frequency wave which is emitted from
the antenna of the transmitting station. Just as the microphone
current is said to be modulated, so is the radio-frequency wave of
the station said to be modulated by these amplified audio-fre-
quency tones.
In spite of the fact that considerable power is used at the
broadcasting station evidenced by the strength of the signals that
leave the antenna, there is an enormous loss in signal strength as we
230 AUDIO AMPLIFIERS
go away from thetransmitter. These radio-frequency waves modu-
lated at audio frequencies are greatly reduced in strength owing to
absorption and dissipation in the medium through which they travel.
It is now our duty tomake up the transmission losses, to amplify them
again, to demodulate or separate the audio- and radio-frequency
currents, and then to forget all about the radio circuit from
then on.
An audio amplifier, then, has nothing to do with radio at all,
and can be used to amplify any voice or music frequency-modu-
lated electric currents that are placed upon its input. Its Job is to
build up the strength of a minute electric current to the point
where the audio modulations on this current are of equal or greater
strength than they were originally.
From the moment the sound in the studio enters the micro-
phone, it ceases to exist as sound, and becomes an electric current.
The steady or d.-c. microphone current, in such a system, is a
carrier for voice or music-frequency modulations, just as the steady
or average d.-c. plate current of a vacuum tube is a carrier for the
a.-c. plate currents that flow there. So too is the radio-frequency
a.-c. current flowing in the transmitting antenna. So too is the
radio-frequency current flowing in one’s receiving antenna and in
the plate circuits of one’s amplifiers. No sound is emitted again
until some translating device is used, a device that will have
electric currents in its input and sound or air waves in its output.
Then, and only then, is sound emitted. Nowhere along the line
from microphone to loud speaker can the signal be heard, unless
some translating device is ‘‘ plugged in.”
The voice-frequency currents that can be amplified by an audio
amplifier may come from a phonograph “ pick-up,” a telephone
transmitter mouthpiece, or the plate circuit of a detector tube in
a radio receiver. The radio link is, then, only an incidental and
extremely inefficient part of the whole system—better results could
be secured by eliminating the radio link completely and using
telephone or power lines between the receiver and the transmitter.
The advantage of radio is the ability to “ broadcast ” in all direc-
tions, and to eliminate the need of a metallic circuit between the
person broadcasting and the person listening.
THE REQUIREMENTS OF AN AUDIO AMPLIFIER 231
Let us forget all about the radio part of our receiver, tempora-
rily, and consider only the audio-frequency amplifier.
192. The requirements of an audio amplifier —An amplifier to
accept, transmit, and amplify audio-frequency tones has several
requirements which have taxed the ingenuity of many engineers:
(1) The amplifier must transmit all the tones required, in their
proper proportion. It must have no frequency distortion.
(2) The amplifier must amplify all frequencies properly whether
the input voltage is high or low. It must have no volume dis-
tortion.
(3) It must have an overall amplification which added to the
radio-frequency amplification will make up for the enormous loss
of power between the input to the microphone and the output from
the loud speaker.
Let us consider, for the moment, only the amount of amplifica-
tion necessary. This value of overall amplification varies, of
course, with the input signals possible, and the output power
required. For home reception an output of one watt of electrical
power into a loud speaker of average efficiency—perhaps 5 per
cent—is sufficient, although many people get along with much less
than this value, and some require much greater power outputs.
If the amplifier works out of the conventional “ grid leak and
condenser ”’ detector tube, the maximum voltage available without
distortion due to detector overloading is about 0.3 volt. If the
loud speaker has a resistance of 4000 ohms and if we fix the output
power at 0.7 watt there must be the following amplification:
W. = power into load = 700 milliwatts
R, = resistance of load = 4000 ohms
E,, = voltage across load Ho = VWo So = 208 VOlts mms:
E = total a.-c. voltage in
plate circuit of last tube = 53 X 3/2 = 79 volts
E; = input voltage to amplifier = 0.3 volt r.m.s.
G = voltage amplification = 79/0.3 = 260 times
It is better that the amplifier shall have considerably more gain
than this figure so that the detector may be worked at a safe dis-
232 AUDIO AMPLIFIERS
tance below its overloading point—to have a good factor of safety
—and that weak signals from distant stations can “ load up ”’ the
amplifier and its speaker.
In some modern receivers certain modifications have been made
so that the detector output is sufficiently high that a loud speaker
can be operated on but one stage of audio amplification. (See
Section 281.)
Let us say, then, that the minimum gain for our amplifier is
300 times. How can we secure this amplification? There are
several ways and we have already discussed the various types of
amplifier arrangements. It is only necessary so to design each
stage or to add a sufficient number of stages that the overall voltage
amplification will be sufficient. The next problem is to see if the
required frequency characteristic can be secured to prevent dis-
tortion due to over emphasis or discrimination at some frequencies.
Example 1-11. An amplifier has a voltage gain, up to the grid of the last
tube, of 200. The last tube has a u of 8 and a plate resistance of 5000 ohms.
It works into an impedance of 10,000 ohms. The input to the amplifier is
0.1 volt r.m.s. Calculate the power into the resistance, the voltage across it
and the ratio between this voltage and the input voltage to get the overall
voltage amplification.
The a.-c, voltage on grid of last tube = H; X G = 0.1 X 200 = 20 volts.
The a.-c. voltage in plate circuit of last tube = 20 & 8 = 160 volts.
The a.-c. voltage across 10,000-ohm load = 2/3 & 160 = 106 volts.
Power into load = H?/R = 1062/10,000 = 1.2 watts.
Voltage amplification = H,/H; = 106/0.1 = 1060 times.
Problem 1-11. What voltage gain must an amplifier have to deliver 1.5
watts to a 4000-ohm load in the plate circuit of a tube whose plate resistance
is 2000 ohms if the input voltage to the amplifier is 0.3 volt r.m.s.?
Problem 2-11. The load of an amplifier is a 2000-ohm resistor. 'The
plate resistance of the last tube is 2000 ohms and its » is 3. The gain in volt-
age up to the grid of this last tube is 100. What input voltage is required to
deliver 50 milliwatts of power?
Problem 3-11. Prove that if the load resistance is double the tube resist-
ance, the voltage across the load is two-thirds of the total a.-c. voltage in the
plate circuit of the tube and the voltage amplification is equal to the amplifica-
up to the grid of the tube times the u of the tube times 2/3.
Problem 4-11. If the load resistance is twice the tube resistance and the
a.-c. voltage across the load is 100 volts, what must be the total a.-c. voltage
in the plate circuit? If the « of the tube is 8, what must be the grid a.-c.
FREQUENCY CHARACTERISTIC OF RESISTANCE AMPLIFIER 233
voltage? If this is the r.m.s. voltage, what must be the minimum C bias of
the tube to keep its grid from going positive?
Problem 5-11. The voltage gain per stage is 8. How many stages will
be needed to produce an overall gain of 500? Remember that voltage gains
are multiplied, not added. Estimate the gain as the total a.-c. voltage in the
plate circuit of the last tube divided by the input voltage.
Problem 6-11. The voltage gain per stage is 10. What is the total voltage
gain (total a.-c. plate voltage divided by input voltage) in three stages?
193. Cascade amplifiers.—It is usually necessary to use ‘more
than one stage of amplification, whether this is at high (radio) fre-
quencies or at low (audio) frequencies. That is, an a.-c. voltage
is applied to the grid of one tube and amplified; this amplified
a.-c. voltage is used to drive the grid of another tube where it is
again amplified. This voltage may again be amplified or used to
drive the grid of a power tube whose function is not voltage ampli-
fication but the delivery of power from the B battery to a loud
speaker. The first tube in a two-stage amplifier is designed to
amplify the voltage alone, and the transfer of power either at
maximum efficiency or at maximum output is not a consideration.
The second tube, or the third in a three-stage amplifier, is designed
solely for the purpose of delivering power to the load, whether this
is a loud speaker, or group of speakers, or a telephone line where it
may be again amplified and put into the loud speakers. When two
or more amplifiers are connected ‘in series” they are said to be
connected in “ cascade.”
MEg,
Fig. 150.—Two tubes coupled by a resistance-capacity unit.
194. Frequency characteristic of resistance amplifier.—Let us
look at the circuit in Fig. 150 which gives two tubes coupled by
234 AUDIO AMPLIFIERS
means of a resistance-capacity unit. Voltage is fed to the grid of
the first tube V71, is amplified there, and reappears across Ro.
This amplified voltage is applied to the grid of the second tube
and amplified, whence it reappears across whatever load is in its
plate circuit. The voltage gain of such a stage is the ratio between
the voltage on the second grid and the voltage on the first grid.
That is,
G=—2 (1)
We have already spoken about the conditions that result in
maximum amplification in such a system, that is, the impedance
in the plate circuit of the first tube must be large compared with
the plate resistance of that tube. That is, the equivalent impe-
dance of Ro shunted by C
and Fk, in series must be
large compared to R, (Fig.
151).
Now the only part of
this circuit which has a
Fig. 151.—Equivalent circuit of Fig. 150. different impedance at dif-
ferent frequencies is the
condenser. To get a good frequency characteristic, then, it is
necessary to choose a value of C which will transmit the lowest
frequency desired, because all other frequencies above this limit
will find less impedance than this value. (The reactance of a
condenser decreases with increase in frequency.) It is necessary
to define the loss which we will tolerate in this condenser.
The voltage that is developed in the plate circuit of VT, is
divided between the plate resistance of that tube and the load
in the plate circuit. The voltage across this impedance is the
output voltage, and is divided between that lost on the con-
denser reactance C and the grid resistance R, of the following
tube. The only useful part is that appearing across R,. The ratio
between the voltage across R, and the voltage across the grid of
tube 1 is
FREQUENCY CHARACTERISTIC OF RESISTANCE AMPLIFIER 235
En, uRoRo
En (R, + Bo) VR? + X24 RoRy
where R,, Ro and R, are in ohms and C is in microfarads.
This ratio represents the efficiency of the whole system from
the grid of tube 1 to the grid of tube 2. Now if the equivalent
resistance in the plate circuit of tube 1 is called R’, and if the
tube’s internal plate resistance is R,, the amplification possible is
Ro
OR + Re io
(2)
and the efficiency of transmission will be
R’o
ReeR (4)
and as described in Section 179 if R’o = 3 R, the voltage amplifica-
tion will be 75 per cent of u of the tube; that is,
All of this output voltage, however, is not impressed on the grid
of the next tube. Some of it is lost in the reactance of the con-
denser. The voltage across the coupling resistance A, then, is
divided between the reactance of the condenser and the resistance
of the grid lead, R,.
Let us assume that the voltage across the coupling resistance
is constant, and see what the relation between the value of C and
the grid leak resistance must be for maximum amplification and
best frequency characteristic.
This a.-c. voltage across the condenser and the grid leak in
series will produce a current
ee eee (5)
he
VOR ae
and the voltage across the grid leak—which is the useful voltage—
will be é
E,, = 1h, (6)
236 AUDIO AMPLIFIERS
Now let us call N, the ratio between H,, and H,, the percentage of
amplification of the full voltage, Hz,, which we want to secure
across the grid leak.
Egy fy
N = a 7
Exe Re exe ve
from which we can get the value of C.
N
ON (8)
DRY (ene
which states that the condenser C must have the value given in
formula 8 if we are to get N per cent of the voltage Ez actually and
usefully impressed across R,.
Now we can see at once that the proper value of C depends upon
two factors: first the frequency and second the resistance of the
grid leak. And so we find that not only does the voltage usefully
impressed depend upon C and &,, but that the frequency charac-
teristic, as well, depends upon these factors. It is not difficult to
see that the frequency characteristic should depend upon a con-
denser, C, but the fact that it also depends upon a resistance—
which in itself has no frequency characteristic—is often overlooked.
For every value of C there is a certain value of R, which will permit
maximum amplification at a definite frequency. The amplifica-
tion will be greater than this value at all frequencies above this
figure.
The relation between these factors is shown in the table below.
N = 90 per cent at N = 80 per cent at N = 70 per cent at
50 cycles 50 cycles 50 cycles
C, mfd. R,, meg. C, mfd. Ry, meg. C, mfd. R,, meg.
0.0132 0.5 0.0084 0.5 0.0062 0.5
0.0066 10) 0.0042 1.0 0.0031 1.0
0.0033 2.0 0.0021 2.0 0.00156 2.0
0.0013 5.0 0.0008 5.0 0.0006 5.0
—
FREQUENCY CHARACTERISTIC OF RESISTANCE AMPLIFIER 237
Calculation will show that when N = 80 per cent at 50 cycles
N = 93 per cent at 100 cycles, and for all values over 100 cycles N
is almost unity—that is, nearly all of the available voltage is use-
fully employed. An interesting relation may be secured from this
formula for C. It is
N
CR SS
; ov 1 Ne 9)
and to attain N = 90 per cent at 50 cycles the capacity of C in
microfarads times the resistance of R, in megohms must be .006.
That is, at 50 cycles a
0.1
0.006-mfd. condenser and
a 1-meg. grid leak will give re rah
90 per cent of the total a x, Pa
voltage available across R 3 x A :
to be usefully applied to % “INXS fem 05
the following tube. Now 8 1 {Tt tofs6
; ; = Ce
if one stage of resistance = ZAN SSO
amplification gives 90 ; pete. “INS
cent of the possible ampli- ,003
fication at a given fre- a
quency, two stages will
give (90 ieee cent)? or 81 Grid Leak Res, Megohms
per cent of the possible Fic. 152.—Relation between grid leak re-
amplification considering gistance, coupling capacity and efficiency
the ratio between the out- (K or N).
put of the second tube and
the input to the first. If the overall amplification of the two
stages is to be 90 per cent of the maximum possible, the value
of N per stage must be the square root of 90 per cent or .95
approximately. The curves in Fig. 152 illustrate this effect.
They were taken from the Proceedings of the Institute of Radio
Engineers, December, 1926, “The Design of Resistance Capacity
Coupled Amplifiers,” by Sylvan Harris.
If, then, we choose the value of R, R,, and C properly we can
attain any desired percentage of the u of the tube at any frequency.
To transmit low frequencies properly requires a larger product of C
—
238 AUDIO AMPLIFIERS
times R,. The question remains, how much amplification can be
attained?
Problem 7-11. Assume the following values and calculate the efficiency:
C =1 mfd.; R, = 500,000 ohms, R, = 200,000 ohms, R, = 60,000 ohms,
f = 800 cycles, and f = 40 cycles.
Problem 8-11. The » of a tube is 30 and its plate resistance is 60,000
ohms. Calculate and plot the voltage amplification as the load resistance
varies from 10,000 ohms to 200,000 ohms.
Problem 9-11. A condenser and a grid leak are in series. C = 0.006
mfd., and R, = 2 megohms. Across them is a 40 cycle-voltage of 10 volts.
Calculate the current flowing, the voltage drop across C and across Ry; and,
assuming the voltage across Ry is the only useful voltage, calculate the effi-
ciency by dividing the voltage across R, by the total voltage and multiplying
by 100 per cent.
Problem 10-11. A frequency of 100 cycles is to be transmitted across the
grid leak and condenser in Fig. 150 at 90 per cent efficiency. If, then, N =
0.9, R, = 500,000 ohms, what must C be in microfarads?
Problem 11-11. Calculate what the product of C X Ry must be for any
given values of V. That is, solve for C < R, in equation (8). Then assume
various values of f from 100 to 10,000 cycles and plot products of C x R, to
give a constant value of NV.
Problem 12-11. What is N if C = 0.006, f = 20 cycles, and Ry = 1
megohm?
195. Overall amplification.—Let us consider a three-stage
amplifier, having resistances, tubes, etc., as shown in Fig. 153.
VT, (#=30) VT, (u=30) VT, (4=3.0)
C; C2
Ro, Rg, Ro, Rg,
+B =O +B =
0.1 Volt 2.25 Volts 50.5 Volts 100 Volts
Fic. 153.—Three stage resistance coupled amplifier.
E,=.1 Volt
The first two tubes have an amplification factor of 30, the
final or power tube has a uw of 3. What is the overall voltage
amplification? If the values of R,C, and R,are properly chosen, the
amplification per stage will be 75 per cent of the u of the tube, and
PLATE BATTERY REQUIREMENTS 239
what the first tube amplifies will be amplified again by the second.
That is, if an input of 0.1 volt is applied to the grid-filament input
of the first tube a voltage of 0.1 X .75 X 30 or 2.25 volts will be ap-
plied to the second, and a voltage of 2.25 X 30 X .75 or 50.5 volts
to the third input; and if the resistance into which this tube works
has twice the value of the plate resistance of the tube, two-thirds
of the total a.-c. plate voltage of the last tube will be applied to
the load. That is, the load will have across it a voltage equal to
50 X 3 X 2 or 100 volts.
The overall voltage amplification in such a case will be 100 +
0.i or 1000. Thisis equal numerically to the product of the voltage
gain of the individual stages. Thus if each stage has a gain
of N, two stages will have a gain of N?, three stages a gain
of N?, etc. In this case it will be (22.5)? X 3 xX 2/3 = 1000
(approximately).
196. Plate battery requirements.—One of the objections to the
resistance amplifier is the excessive plate battery voltages that are
necessary. The voltage on the plate of a tube in whose plate cir-
cuit is a high resistance is not the voltage of the battery but this
voltage minus the voltage drop across the resistor in the plate
circuit. We want the a.-c. voltage drop across this resistance to
be high but the d.-c. voltage drop to be low. As an example sup-
pose a tube has a plate resistance, R,, of 60,000 ohms when the
plate voltage (H,) is 100 volts. To get 75 per cent of the u of
the tube as the overall amplification from grid input voltage to
the voltage output across the resistor, we need a resistance in the
plate circuit of 180,000 ohms. Suppose under these conditions the
plate current is 1 ma. What plate battery voltage is necessary if
100 volts are required on the plate?
The plate voltage may be found from
Ep = by — 1h,
100 = EF, — 0.001 X 180,000
E, = 280 volts.
l|
Now the tube will amplify with fewer volts on the plate than
this—as many manufacturers and experimenters have proved—but
240 AUDIO AMPLIFIERS
the difficulty lies in the following fact. The plate resistance of a
tube is a function of the plate voltage. That is, at low plate volt-
ages the plate resistance is high, which in turn necessitates a higher
value of load resistance to get the 75 per cent of the u of the tube,
which means that the voltage drop will be high across this resistor
and so on. A high plate battery is always needed.
To get 100 volts on the plate of this tube requires a plate battery
of 280 volts. This is inefficiency, of course. What can be done
about it?
197. Inductance load amplifier.—Suppose instead of the resis-
tor in the plate circuit we substitute a low resistance inductance
with a high value of reactance
at the frequencies at which the |.
amplifier is to work, for example |
a 100-henry choke coil with a d.-e. b
resistance of 1000 ohms. The cir- e L&R
cuit looks like Fig. 154. Now the L:
d.-c. plate current encounters II I
no appreciable opposition in the Fie. 154.—Inductance-capacity or
1000-ohm resistance. In fact “impedance”’ coupled amplifier.
the voltage drop is 1 volt per
milliampere of plate current. The a.-c. current, however, must
flow through the high inductance and the resistance in series.
Across this impedance (V R? + L?w*) will appear the amplified
a.-c. voltage which may be used to drive another amplifier stage
if desired.
Such an amplifier is commonly called an impedance amplifier.
The loss in d.-c. voltage between the plate battery and the plate
itself is only 1 volt per milliampere current, and if the plate
current is 2 ma. only 102 volts of B battery will be required to put
100 volts on the plate of the tube.
Problem 13-11. An amplifier is to be used at audio frequencies. What
must be the impedance of the choke coil at 100 cycles to secure an amplifica-
tion of 7.5 from a tube whose yu is 10 and whose plate resistance is 15,000 ohms?
What will be the amplification at 1000 cycles?
Problem 14-11. If the resistance of the choke coil used in Problem 13 is
500 ohms, what is the inductance required?
EFFECT OF STRAY CAPACITIES AT HIGH FREQUENCIES 241
Problem 15-11. A tube has a load resistance of 100,000 ohms. The plate
current is 2 ma. and the plate voltage required is 90 volts. What plate battery
voltage is necessary?
Problem 16-11. A tube draws 0.5 ma. from the B batteries whose voltage
is 180 volts. A coupling resistance of 100,000 ohms is used. What is the
voltage actually on the plate?
198. Effect of stray capacities at high frequencies.—When one
sets up a resistance amplifier in the laboratory and with a constant
input voltage at various frequencies measures the voltage, the
slight falling off at the lower frequencies is noted, but if the
capacity of the coupling condenser and the grid leak are properly
proportioned the droop at 100 cycles or below will not be pro-
nounced. But the laboratorician will probably note a falling
off at frequencies beyond 5000 cycles—which was not foretold
by any of our mathematics or analysis up to the present moment.
What is happening to the higher frequencies?
Let us look at Figs. 155 and 156 in which several additional
capacities appear. These are made up of the internal plate-to-
Cop
Coy
Cor
Fic. 155.—Equivalent circuit at high Fic. 156.—Tube capac-
frequencies of resistance-capacity coupled ities.
tubes.
filament capacity of tube 1, the combined plate-to-filament and
grid-to-filament capacity of the tube 2, and capacities in the
sockets, wiring, etc. All of these capacities are directly across
the grid-leak resistance or the input to the second tube. Their
effect is small at low frequencies because they are small capacities,
and have high reactances at low frequencies. They are, then,
bridging impedances of very high value and consequently have
little effect at low frequencies. But at high frequencies their
242 AUDIO AMPLIFIERS
impedance decreases, and by-passes some of the high-frequency
voltages that normally appear across the grid leak. The higher
the frequency, and the greater the effect of these additional and
unavoidable capacities, the greater the droop of the curve repre-
senting the response of the amplifier plotted against frequency.
The droop at high frequencies depends too upon the overall gain
of the amplifier; therefore when we boost the gain by using a large
coupling resistance, R,, we do get more amplification in the middle
of the audio-frequency band, but a more decided drop at both low
and higher frequencies.
These effects are illustrated in the curve in Fig. 157, taken from
a Radio Club of America paper
P L=300 Henri by A. V. Loughran, published in
3 Radio Broadcast, August, 1927.
os Rp=1,000,000 Ohms The small capacity existing
< Bae between the filament and the
= I |Rp=250,000 Ohms : ;
5 | | grid, etc., is due to the fact that
i o(=0.005 Mi, they are made up of conductors
es 8s 8 8s 8 = insulated from each other; our
Frequetey-Cycles 4~=S(“‘“ e:«C*#«nition of a capacity. They
Fic. 157—Frequency characteristic @re at different potentials. In
of resistance- and inductance-coupled addition there are capacities exist-
amplifiers. ing between the terminals on the
tube socket, and in the wiring of
the amplifier. Allin all, these unwanted and undesirable capacities
may ruin an otherwise perfect amplifier—and they must always be
reckoned with.
Not only do these capacities, which are represented in Fig. 155,
play havoc with high frequencies, but the effect of the grid-plate
capacity of the second tube is multiplied by the u of the tube—a
very interesting and unfortunate fact.
The capacities which cause trouble at high frequencies are:
(1) Grid-filament capacity, Ce
(2) Plate-filament capacity, C,,,.
(3) Plate-grid capacity, C,,.
(4) Stray capacities in wiring, etc.
EFFECT OF STRAY CAPACITIES AT HIGH FREQUENCIES 243
The capacity C, across the input to the tube is equal to
Cy = Coy — Co» (uo 1),
where ju is the effective amplification in the circuit, and the other
factors are as itemized above and shown in Fig. 155.
These values for the Radiotron UX-240, a tube adapted for
resistance-coupled amplifiers, are Cp, = 1.5 mmfd., Cy, = 3 to
4 mmfd., and C,, = 8.8 mmfd. In practice C, may vary from
20 to as high as 300 mmfd. depending upon the intrinsic values of
its components and the amplification of the system. That is, if
the Cy, = 3.0 mmfd., and C,, = 8.0 mmfd. and the effective ampli-
fication of the system is 20, C, becomes 3 plus 8 X 20 or 163 mmfd.
—which is an appreciable capacity and not at all what one would
expect. The 201—-A has about the same inter-electrode capacity
and the UX-171 has somewhat greater grid-filament capacity
owing to the longer electron-emitting surface. The effective capac-
ity across the input to the tube is a function of the uw of the tube;
thus the greater the amplification factor the more trouble one gets
into because of this capacity-multiplying effect. Low u tubes are
not so troubled, but their amplification is low; and so the amplifier
designer and experimenter must compromise between a good fre-
quency characteristic and a high gain. If the gain is good the high
frequencies are discriminated against, and if the characteristic is
very good the overall gain is likely to be low.
Problem 17-11. The pu of a tube is 8, its grid-filament capacity is 6 mmfd.,
its plate-grid capacity is 8.0 mmfd. The effective amplification of the circuit
is 7, and stray capacities across the filament and grid circuit amount to 2 mmfd.
What is the effective shunting capacity Cy?
199. Quantitative effect of capacities on high frequencies.—
How much do these shunting capacities cut down the high fre-~
quencies? The effective resistance of a resistance shunted by a
condenser (as C, and R, in Fig. 155) is:
ert
~ 1+ a RC”
r
and in the table below may be found some values of the effective
.
244 AUDIO AMPLIFIERS
resistance of a quarter-megohm resistor shunted by 60 mmfd.
capacity.
Frequency Effective resistance (ohms)
2,000 240,000
4,000 220,000
6,000 190,000
8,000 160,000
10,000 130,000
If the resistor is 1 megohm and if the shunting capacity is 300
mmfd., the effective resistance will be only 2800 ohms at 10,000
cycles although it is 960,000 ohms at 100 cycles—a drop of 97 per
cent.
These figures show that the higher the plate-coupling resistance
the greater will be the discrepancy between the amplification at
10,000 and 100 cycles. This is because the reactance of the
capacity is the same at 10,000 cycles whether it is across 1 meg-
ohm or across 7g megohm, but the shunting effect in the first
case is much greater than in the second. Thus if 10,000 ohms
is shunted across 1 megohm, the resultant decrease in impedance
will be from one million ohms to not far from ten thousand ohms
or a reduction of one hundred times. If, however, it is shunted
across 100,000 ohms it has decreased the impedance only ten to one.
Example 2-11. Suppose a 1-megohm resistor is in the plate circuit of a
tube and that 10 volts are developed across it. If 100,000 ohms are shunted
across it, the effective impedance becomes 0.91 x 10°, and if the current is
the same the voltage developed has been reduced by 91 per cent. Now, sup-
pose 10 volts are developed across 100,000 ohms and that another 100,000-ohm
resistor is shunted across it. What is the resultant reduction in voltage?
Clearly it is 50 per cent—a much smaller percentage reduction.
Problem 18-11. A tube with an amplification factor of 30 and a plate
resistance of 150,000 ohms is used with a half-megohm coupling resistance.
Suppose that in construction a path of soldering flux is placed across the
terminals of this resistor accidentally so that the half-megohm is effectively
shunted by 100,000 ohms. What is the resultant amplification? What is
the amplification without the shunting resistance? What is the percentage
loss due to the flux?
These capacities within the tube and those due to the socket
and the wiring reduce the amplification at the higher audio fre-
quencies and make impossible a flat characteristic with high over-
TUNED INDUCTANCE AMPLIFIER 245
all gain. If one stage loses 50 per cent of the response at 5000
cycles, two of them will lose 75 per cent and three of them 87.5 per
cent, and so on.
The curve in Fig. 157 showing the recommended values for use
with the UX—240 tube is a very good characteristic, and equals
many and betters some of the amplifier characteristics obtained
by other coupling devices.
200. High-frequency response in impedance-coupled ampli-
fiers.—The main disadvantage of the resistance amplifier, that is,
large B battery voltages, may be overcome by using high induc-
tance chokes instead of the resistor, but the loss at high frequencies
is even worse because of the distributed capacity of the winding
of the choke. A characteristic using 300-henry chokes appears in
Fig. 157 and as may be seen is worse than the resistance-coupled
stages. It is possible to design a better choke than was used in
this experiment, but even then great care must be taken to preserve
a good frequency characteristic.
201. Tuned inductance amplifier.—In both the resistance load
and the inductance load amplifier the maximum amplification is
attained when the impedance in the plate circuit is a maximum.
If, then, we desire to transmit only one frequency, say 1000 cycles,
we can get greater amplification by
tuning the inductance with a shunt
condenser so that we have an anti-
resonant circuit in the plate circuit
of the tube, as shown in Fig. 158. If I
an inductance of 0.1 henry is tuned Fre. 158.—Tuned inductance
with a 0.254-mfd. condenser the output.
resonant frequency will be about 1000
cycles. If the coil has a resistance of 100 ohms the impedance of
the anti-resonant circuit will be roughly 3950 ohms whereas the
reactance of the coil alone will be only 628 ohms. Tuning the
coil therefore increases the load impedance in the plate circuit by
about 6.3 times. ;
The voltage gain will be given by the formula of Section 178:
— PPeLias'
ee een.
G
246 AUDIO AMPLIFIERS
which shows that the maximum possible amplification is the » of
the tube, and, as we have already seen, increasing the load impe-
dance R, to the point where it is three times the tube internal
resistance results in a voltage amplification of u X .75.
Problem 19-11. An inductance of 170 microhenries has a resistance of
6.0 ohms at 500 ke., 10 ohms at 1000 ke., and 18 ohms at 1500 ke. Assuming
a fixed capacity—due to distributed capacity of winding, connections, etc.—
across the coil of 60 mmfd., calculate the condenser capacity that will tune
the coil over the broadcast frequency band. Calculate the reactance of the
coil at the three frequencies above and the voltage gain to be expected when
used with a tube whose plate resistance is 12,000 ohms and a yp of 8. Then
calculate the impedance in the plate circuit if the inductance is tuned at each
of the three frequencies and the voltage gain to be expected when connected
as in Fig. 158. Plot the voltage amplification of this single stage against
frequency. Explain why the curve is not flat.
The untuned inductance or ‘‘ impedance” amplifier can be
used where it is desired to transmit a fairly wide band of fre-
quencies; tuning it makes it possible to get greater amplification
over a narrow band of frequencies.
202. The transformer-coupled amplifier.—The output a.-c.
voltage across the resistance or inductance—tuned or not—can
never be higher than the input grid voltage multiplied by the u of
the tube, and can attain that value only when the resistance or
impedance is high compared to
the plate resistance. Suppose,
Ws, P however, we use a transformer,
Ey |Re Ep as in Fig. 159. The voltage
= across the secondary will be
! ie L increased by the turns ratio of
Fre. 159.—Transformer coupled am- the windings and so the voltage
plifier developed in the plate circuit
of the tube may not only be
passed on to a following tube multiplied by the u of this tube
but multiplied by the turns ratio as well.
If the secondary circuit takes no current or power the greatest
voltage will appear across the secondary when a very high turns
ratio is used, but this is not the case when power is taken—and
some always is.
a THE TRANSFORMER-COUPLED AMPLIFIER 247
The maximum voltage across the secondary will be obtained
when the turns ratio is given by the expression
where F, and R, are the resistances between which the transformer
works. When such a turns ratio is used the voltage across the
secondary is given by
M1 E,,N
He (max, ) = ore (7)
All of this assumes that a perfect transformer is used, that is,
one which has no d.-c. resistance, no magnetic leakage, and
infinite primary and secondary reactances. If the resistance of the
load across the secondary is 1 megohm, and the plate resistance
of the tube is 10,000 ohms, the proper turns ratio,
/l 000,000
= ea ee 1 ()
M 10,000 ,
and the voltage gain from (7) is
E,
Ey,
=m X5 = 40if m = 8.
The foregoing mathematics discloses several interesting points. In
the first place it is possible to get much greater voltage amplifica-
tion by means of a transformer than is possible with the same tube
and either resistance or inductance output. In the second place,
for every ratio of resistance across the secondary and primary of
the transformer there is a certain turns ratio which will produce
the maximum voltage step-up. This means that once the resist-
ances on either side of the transformer are determined, the turns
ratio for maximum voltage gain is fixed. This, in turn, means that
248 AUDIO AMPLIFIERS
a transformer can be used as an impedance adjusting device, that
is, a coupling device between two circuits of different impedance,
one of which acts as a source of voltage and the other is the recipi-
ent of a voltage, amplified or not.
Whenever the impedances of the two sides of the transformer
are not that indicated by the expression above, there is a loss in
secondary voltage, a transmission loss, engineers say. Whenever
two circuits are to be coupled together with the least loss in voltage
or power the proper turns ratio transformer must be used, that is,
N? = Z,/Z,, where Z, and Z, are the impedances between which
the transformer works.
Problem 20-11. <A tube whose plate resistance is 12,000 ohms (UX—201—A)
works into a resistance, f,, of 600,000 ohms. What is the proper turns ratio,
and what is voltage gain if » = 8? If #H,, = 1 volt what voltage will appear
across R,?
Problem 21-11. One tube whose plate resistance is 25,000 ohms is con-
nected to another whose grid circuit has a resistance of 400,000 ohms. What
is the turns ratio of the transformer for maximum voltage amplification?
Problem 22-11. An “output” transformer is to be used to connect a
loud speaker to a tube. The impedance of the loud speaker at the desired
frequency is 4000 ohms, the tube has a resistance of 2000 ohms. What is the
proper value of NV?
Problem 23-11. A telephone line has an impedance of 600 ohms. The
a.-c. voltages in the plate circuit of a 6000-ohm tube are to be transferred to
this line. What must be done to effect such a transfer with least power loss?
203. Transformer with no secondary load.—If the secondary
of the transformer works into a true no-load impedance, that is, an
open circuit, or A, = infinity, the voltage appearing across the
secondary will be wH, multiplied by the turns ratio of the trans-
former and not, as given in (7) wH, X N/2. The voltage gain is
equal to
Nu
Coa 2 xfliN a Ne Rp \2
VRP + 2rfLy2 NIT ( a)
where L; = inductance of the primary, etc.
Example 3-11. Transformers of a few years ago had very little primary
inductance. Assume an inductance of 2 henrys, f = 800 cycles, 1 = 8, and
THE ADVANTAGE OF THE TRANSFORMER 249
Rp = 10,000 ohms, N = 4. Calculate the voltage amplification. Then
assume f = 80 cycles and calculate G.
ee 2x 3.14 x 800 X 2x 4
V108 + (2 X 3.14 X 800 X 2)
= 22.7
If f = 80 cycles
z 2x 3.14 x 80X24
V108 + (2 x 3.14 X 80 X 2)?
Problem 24-11. Assume a primary inductance of 50 henrys and calculate
the gain per stage G using the other values used in Example 3. Assume f =
80, 800, and 8000 cycles and plot the characteristic, that is, gain against
frequency.
204. The advantage of the transformer.—The tranformer has
the great advantage that it can contribute toward the voltage
amplification, and can contribute toward the maximum power
output when the load resistance or impedance differs from the tube
resistance. In addition, high B battery voltages are not necessary.
A transformer can be constructed so that it has little loss in
itself. This loss is due the d.-c. resistance of the windings, the fact
that perfect coupling between primary and secondary is not
attained, and because of iron losses, that is, currents set up in the
iron core represent a loss in power which must be supplied from
the source. This power therefore does not get out of the trans-
former and cannot appear in the output.
If a good transformer is used, its transmission loss is small. Its
effect, then, upon the circuit is of two sorts; first it may contribute
toward the voltage gain by having a voltage step up in it; second,
it may be used to ‘‘match”’ the load resistance to the tube resist-
ance. Let us suppose the tube resistance is 10,000 ohms, and that
the secondary load resistance is 100,000 ohms. A resistance of this
value interposed in the plate circuit of a tube will not have the
maximum power developed in it. (See Section 180.) But if we
use the proper transformer such that N? = 10, this 100,000 ohms
will look, to the tube, like a resistance of 100,000 X .01 or 10,000
ohms—the condition for maximum voltage and maximum power.
The transformer can always be forgotten if we substitute for
250 AUDIO AMPLIFIERS
it and its secondary load this load divided by N?. Looked at from
the secondary we may replace the transformer by N? X Zp.
Example 4-11. A transformer with a turns ratio of 3 connects a 10,000-
ohm tube with a load which has a frequency characteristic such that at 100
cycles the load has one-tenth of the impedance it has at 1000 cycles. The
load impedance at 100 cycles is 90,000 ohms. ‘The yu of the tube is 8. What
are the voltage amplification and the power output at 100 and 1000 cycles?
The transformer may be dispensed with in the calculation if we transfer the
secondary load directly into the plate circuit of the tube by multiplying
it by 1/N2, that is, 1/9.
Thus at 100 cycles the impedance in the plate circuit is
90,000 x 4 = 10,000 ohms
and the voltage amplification is
uX Ro 8 X 10,000
Re-aR,. 620,000.
and the power
_ nH? Ro
~ (Ro + Ry)?
= E,? X 1600 x 10-*
Ie,
Ss
at 1000 cycles, 18.
900,000 x 4 = 100,000 ohms
8 X 100,000
d G = ————_ = 7: » = E,? —6
an 110,000 7.3 and P, = E,? X 533 X10
Problem 25-11. Assuming no loss in the transformer, what is the power
transmitted to a 6000-ohm load from a 2000-ohm tube when they are coupled
by a transformer whose secondary winding has 2.24 times as many turns as
the primary? Assume » = 3, EH, = 10 volts r.m.s. Use formula (5) in Section
180. What would be the value of N for maximum power in the load?
What would be the power then? What value of N would deliver maximum
undistorted power output into the load? What is this value of power in
milliwatts?
205. Effect of leaky condenser in resistance-coupled amplifier.
—The coupling condenser C (Fig. 150) must have a very high
d.-c. resistance, the resistors R, and R, must be quiet even though
appreciable current goes through R, and high a.-c. voltages appear
across k,. Suppose the preceding tube has a d.-c. plate resistance,
LEAKY CONDENSER IN RESISTANCE-COUPLED AMPLIFIER 251
Rf, of 20,000 ohms under the conditions of operation, that is, 90 volts
B battery, a coupling resistor of 60,000 ohms, and a plate current
of 1 milliampere. Suppose the condenser has a d.-c. resistance of
10 megohms (10 million ohms) and that the grid leak of the fol-
lowing tube is 1 megohm. What happens?
Across the 60,000-ohm resistor is a 60-volt d.-c. voltage drop.
This is impressed upon the two impedances, C and R, in series,
11 megohms. By Ohm’s law the current that will flow in this
circuit is 5.45 microamperes, which across the 1-megohm grid
leak produces a d.-c. voltage of 5.45 volts. This voltage is of such
a polarity that the grid end of the grid leak is positive 5.45 volts.
Suppose that the C bias is 4.5 volts and that a maximum or peak
input voltage on the grid of this tube of 4.0 volts is expected. The
actual C bias is 5.45 — 4.5 or 0.95 volts and so the tube is in
imminent danger of going positive at any instant.
Let us suppose that the grid goes positive on a strong signal.
It draws electrons from the stream leaving the filament and going
toward the plate. These electrons make the grid negative, and if a
sufficient number of electrons is trapped on it by a large positive
overload the grid is so negative that the plate current is reduced
to a very low value or even zero.
_ The tube now begins to “ block,” that is, the electrons may leak
off to the filament at an audible rate and the tube emits “ putt-
putting ” sounds which entirely destroy reception.
The grid of a resistance-coupled amplifier must never be allowed
to become strongly positive, either by the introduction of too
strong signals or through an improper C bias due to leakage of the
B battery current voltage through the coupling resistor. If the
grid is insufficiently biased and takes a momentary supply of elec-
trons, these flow back to the filament through the grid leak and
form a negative bias across this resistance. This bias aids the
steady bias from the C battery and thus in a way tube overloading
is self-correcting.
The condenser and grid leak are used in the inductance-coupled
amplifier. If the resistance grid leak is replaced by a high-reac-
tance, low-resistance choke, “‘ blocking ” does not occur so readily
because the electrons leak off more rapidly.
¥
252 AUDIO AMPLIFIERS
206. Reflex amplifiers—When two sets of frequencies of
widely different order are to be amplified it is possible to make
one tube serve as both a low-frequency and as a high-frequency
amplifier. Such systems in which the energy from a tube is
fed back into the input of a previous tube in a cascade amplifier
is called a refiex system; although such systems are not in
vogue at the present time they were of very great importance
and interest in the early days of broadcasting. Let us look at
Fig. 160.
Radio-frequency signals are impressed across the coupling coil
VT
[" eee
-001
&
oLeLeKororere
001
P
les
Fic. 160.—A reflex circuit. V7 amplifies both at low and high frequencies.
L, and from thence are transferred to Lz where the required signals
are selected by the tuned circuit L2C;. These voltages are ampli-
fied in the tube V7’, and after a further selection process goes on
in the tuned input, L4C2, are passed on to the tube VT, which is a
detector. In the plate circuit of this detector tube the audio fre-
quencies appear. They are passed back into the audio transformer
secondary, S, which is in series with the grid of the VT,. There is
no voltage loss at audio frequencies in Le because it has practically
no impedance at these frequencies compared to the impedance of
the grid-filament path of the tube or of the transformer. In
TRANSFORMER-COUPLED AMPLIFIERS 253
this tube the audio frequencies are amplified and are passed into
the head phones or loud speaker or other audio amplifiers as noted
at L.S. This load is by-passed by a large condenser so that no
radio-frequency voltages are lost there. Of course there are audio-
frequency voltages across the primary, L3, of the inter-tube trans-
former but they are very small because this primary has practically
no impedance to audio frequencies. And because of the very poor
coupling to tube 2 at audio frequencies, practically no audio volt-
ages appear across the input to the demodulator tube VT,. We
have then two tubes serving as radio-frequency amplifier, detector
and low-frequency amplifier.
207. The inverse duplex.—The first tube in a radio set is the
one which has the least work to do—the incoming signals are at
their lowest level. The grid voltages are lowest in this tube. The
next tube has greater signals to handle, and the final tube in a
receiver must handle the greatest grid voltage swings. In the
Grimes inverse duplex system, the first tube handles not only the
low radio-frequency signal voltages but the strongest audio-fre-
quency voltages, the second handles moderate radio-frequency and
moderate audio-frequency voltages—and so no tube is overloaded
with its double duty.
So long as the different sets of frequencies are far enough apart,
the reflex amplifier works fairly well, but the difficulties of filtering
together with poorer stability in such amplifiers militate against
their wide use. In addition, tubes are not so expensive now but
that the average experimenter prefers stability to a slight and
questionable economy.
208. Transformer-coupled amplifiers.—The single transformer
has already been discussed. It was stated in Section 202 that the
maximum ratio for a transformer working between a 10,000-ohm
tube and the input of another tube which might have an impe-
dance of 1 megohm was 10 to 1. Unfortunately, this is only
theoretically true, just as the calculations on resistance coupling
without regard to certain factors produce erroneous results. A
transformer to give a good low-frequency response when worked
out of a detector tube—which normally has a rather high plate
254 AUDIO AMPLIFIERS
resistance—must have a primary inductance of about 100 henrys.
Now a 9 to 1 transformer would have a secondary inductance of
9X9 100 or 8100 henrys, and unfortunately such a trans-
former cannot be wound without its secondary having considerable
capacity between layers of wire and between individual turns.
This secondary distributed capacity shunts out the high frequencies
just as the stray capacities in a resistance- or inductance-coupled
amplifier lose the high audio tones.
It has not been found possible to build a transformer which
will yield a flat frequency characteristic when worked out of a high
impedance tube, and when using ordinary iron for the transformer
core, with a turns ratio of much greater than 3 to 1. Using higher
permeability iron, a somewhat greater turns ratio can be secured
because less wire need be used to get a given value of inductance
but there are very few transformers on the open market with a
turns ratio of more than 4 to 1 that give a flat charac-
teristic.
This statement does not preclude the possibility of using a
higher ratio and overcoming the loss of either low or high tones—
or both—in some other part of the circuit, and in fact many
amplifiers have been so designed.
Figure 161 is a characteristic of a single transformer obtained
by putting known voltages across the primary and measuring
the secondary output voltage.
The hump at 5000 eycles is due the leakage inductance between
primary and secondary resonating with the secondary distributed
capacity. Beyond this point the whole transformer looks like a
capacity to the tube and hence the effective resistance into which
the tube works becomes steadily less as the frequency in-
creases.
209. Measurements on transformer-coupled amplifiers.—The
curve in Fig. 161 on a single transformer may be no indication at
all of what a two-stage amplifier may do. This is due to the
fact that a certain amount of “regeneration” takes place in
the average amplifier unless considerable pains are taken to
prevent such difficulties. This regeneration distorts the curve,
and makes laboratory measurements impossible to check. One
CALCULATION OF OVERALL VOLTAGE AMPLIFICATION 255
day the curve may be one thing, and on the next it may be
different.
W
|
&
i)
>
10,000 Ohms
re vt
100 200 500 1,000 5,000 10,000
Cycles
Fra. 161.—Characteristic of single andio transformer.
210. Calculation of overall voltage amplification.—Let us con-
sider the amplifier in Fig. 162. The overall amplification is the
H=8 1:3 =8 iil
+B
r +B lt
Fic. 162.—How a voltage EH; is amplified in a transformer coupled amplifier.
ratio between the voltage, EH. across the output load and that
across the input, E;.
256 AUDIO AMPLIFIERS
EEX BGK Ro
etre Ej Ro + Ry
_ . 216 Re
SUR ACR, ake
where RF, = plate resistance of last tube.
It is actually not possible to realize all this amplification
because it is never possible to realize the full » of the tubes.
The voltage gain measured and calculated will check very closely,
usually.
Now suppose we apply an input voltage of, say, 0.1 volt across
the primary of the first transformer. Across the secondary this
will become 3 X 0.1 or 0.3 volt and so the value of the C bias for
this first tube which is fed out of the secondary of 7; need not be
greater than 1 volt since that will take care of severe overload, and
if the tube is a 201—A type a plate voltage of 45 will be sufficient.
Somewhat better amplification and fidelity will result by using
90 volts on the plate and 4.5 volts bias on the grid, since the plate
resistance will be somewhat less under these conditions. If the
impedance of the following transformer as looked at from the tube
is high compared to the FR, of this first tube—as it will be at fre-
quencies of the order of 1000 cycles if the transformer is any good
at all—nearly all of the uw of this tube will be realized. Across the
primary of the second transformer, 72, then will appear a voltage
of 0.3 X 8 or 2.4 volts and across the secondary a voltage of 7.2
volts which will require C bias of 9 volts and about 135 volts on
the plate. The tube should be a 112 type. This voltage will be
multiplied by 8 times in this tube and will appear across the output
load as 35 volts provided the load is twice the resistance of the last
tube. This voltage across a 10,000-ohm load will produce 122
milliwatts of power.
Now let us consider the combination of resistance and trans-
former coupling as shown in Fig. 163. In the constructor’s mind
there exists some doubt as to whether he should use the transformer
next to the input or next to the power tube. Starting with an
“EQUALIZING”’ 257
input of 0.1 volt the a.-c. voltages at various points in the circuit
are as indicated. The second tube which has a resistance input
must have a C bias of more than 2.4 volts, which means that the
plate voltage must be rather high. If, on the other hand, the two
resistance stages precede the transformer-coupled stage the C
bias on the two tubes need not be very high, the plate resistance
will not be so great, the coupling resistor need not have such
a high resistance, and the B battery voltage can be lower in
value.
Other combinations of transformer-, resistance-capacity, etc.,
amplifiers may be and are used.
1:3 -=8 -=8 f=3 1:1
PU
Fra. 163.—A combination transformer and resistance coupled amplifier.
211. ‘‘ Equalizing.”—It is possible by means of correcting
circuits to get almost any kind of frequency response curve desired.
For example, if a certain amplifier is deficient at low frequencies,
one stage may be made resonant to these frequencies and so pull
up the characteristic. If the amplifier tends to sing at some
frequency, a loss may be put in at this frequency, and the overall
curve will be flatter than without the equalizer. In the telephone
plant the use of equalizing networks is very important and has
come to be almost an exact science. These consist of resistances,
inductances, and condensers. Equalizing usually results in an
overall loss in amplification; that is, some loss is incurred which
must be made up by additional stages. In other words one gains
a better characteristic at the expense of amplification. In some
amplifiers a better low-frequency response is secured by tuning the
primary of the transformer by a capacity to a low frequency.
258 AUDIO AMPLIFIERS
212. The power amplifier.—The final tube in an audio amplifier
which is feeding audio frequencies into a loud speaker must be
essentially a power amplifier. Its task is to deliver undistorted
power to the loud speaker, and not to develop any great amount
of voltage amplification. The task of the previous amplifier
stages is to build up the small output voltages of the detector so
that the large voltages necessary to swing the grid of the power tube
may be obtained. The a.-c. power in the plate circuits of tubes
previous to the power stage is small; what is required is that each
previous stage shall give a maximum of voltage amplification with-
out distortion, and the fact that maximum power may not be
developed in these plate circuits is not important. These tubes
work into very high impedances in which it is not possible to
generate much power although it is possible to build up consider-
able voltages across them.
The a.-c. plate current of the last tube, then, must be rather
large and this means that the grid a.-c. voltages must be large,
which in turn means that the H,—I, curve of this tube must have
a long and straight part. The 171 tube, for example, which can
deliver about 700 milliwatts without much distortion must have
an r.m.s. grid voltage applied to it of about 27 volts; there must
be a portion of the #,—I, curve which is straight over at least twice
this number of volts. Thus if the grid is biased 40.5 volts, the
characteristic must be straight from minus 78.5 to minus 2.5 volts.
The next preceding tube has much smaller voltage swings to handle
and so its characteristic need not be straight over such a long part.
Because of the high cost of power apparatus when power tubes
requiring high voltages are used, the final tube in the average radio
receiver is a low yu, low-resistance tube which will deliver consider-
able power to the loud speaker without demanding excessive plate
voltages. Such tubes require large input a.-c. voltages and there-
fore require greater amplification before them than tubes of higher
values of u. This can be looked at in another way. Modern
receivers have considerable amplification, and, to prevent any
danger of overloading power tubes with consequent distortion, it
is necessary to use tubes with considerable C bias and capable of
THE PUSH-PULL AMPLIFIER 259
handling large swings of voltage. The power tube of the 171, 245,
or 250 class is a tube of that type. Such tubes will handle more
input voltage than a high-y, higher-resistance tube; they require
more input voltage to deliver a given amount of power.
Distortion with such tubes is not due so much to the grid’s
going positive on strong signals as it is to the curvature of the
characteristic at the lower end. The grid is not forced too positive
but too negative, and so the negative and positive halves of the
a.-c. cycle are not amplified symmetrically.
The table below shows the advantage of the 245 type of tube,
delivering considerable power with a moderate plate voltage.
Power Output,
Types of Tube Plate Volts
Watts
UX 171-A OZ 180
UX 210 def 425
UX 250 4.6 450
UX 245 16 250
213. The push-pull amplifier—The push-pull amplifier de-
creases distortion caused by working a tube over a curved part of
its characteristic. The curves in Fig. 164 show how the plate
current curve flattens out as the load resistance in a 210 type of
tube is increased.
If the current increases and decreases equal amounts above
and below its average value determined by the C bias, all is well
and good. There is no second harmonic distortion. But let us
look at the curves. With the 1000-ohm load the increase is 14 ma.
and the decrease is 11 ma. With a 5000-ohm load the increase
and decrease are respectively 7 and 6 ma. With a 10,000-ohm
load the increase and decrease are 5 ma. each. ‘These figures
indicate clearly that as the load becomes larger the inequalities
between the positive and negative halves of the plate current
cycle become less.
260 AUDIO AMPLIFIERS
These inequalities signify that harmonics are being generated
and that if a pure 1000-cycle voltage is put on the grid, the plate
alae
Signal Voltage
N
60 750 40 30 20 10
E Z Negative
Fic. 164.—Flattening effect of increasing load resistance.
circuit will contain not only a pure 1000-cycle tone but a 2000-cycle
tone as well. Distortion is being produced.
=a
+ 3 —>E,
Fic. 165.—Push-pull amplifier.
The push-pull amplifier circuit is shown in Fig. 165. It consists
of a center-tapped input transformer, two tubes of identical charac-
teristics, and a transformer, or a choke, with a center tap. When a
voltage is induced into the secondary winding of the input trans-
THE PUSH-PULL AMPLIFIER 261
former one grid becomes positive a certain amount and the other
grid becomes negative the same amount.
If the C bias, which is attached to the center of the input wind-
ing, is 35 volts and if across the entire secondary winding appears
a peak a.-c. voltage of 20, one tube has its grid voltage increased
by 10 volts, or to 45 volts, and the other decreased by 10 volts, or
to 25 volts. Thus one plate current increases and the other
decreases. These variations in a.-c. plate current flow through the
output winding and may be transferred to the load. These plate
circuit current variations are out of phase by 180°, one increasing,
the other decreasing a like amount. One tube ‘ pushes ”’ current
through the output, the other “ pulls’ current through it. Sup-
pose both tubes pushed at the same time. What would happen?
If they both pushed the same amount and at the same time no
current would flow through the output winding because the two
currents would neutralize each other.
If, then, we can cause the second harmonic currents to be in
phase, that is, to push or pull at the same instant, and the same
amount, they will not get into the load, and distortion due to these
extra currents will not appear in our loud speaker. This is exactly
what the push-pull system does.
Because the input voltage across such an amplifier is divided
into two parts this amplifier requires twice the input voltage to
give the same output power—unless the turns ratio of the input
transformer is doubled, which is difficult to carry out in practice.
Because of the push-pull arrangement, however, considerable over-
loading can be tolerated before the third harmonics which are not
canceled out become objectionable. And there is another advan-
tage—alternating current can be used to light the filaments of the
tubes, with greater freedom from hum. The output transformer
or choke need not have such a large core because of the fact that the
d.-c. currents in the two halves are flowing in opposite directions;
and since the two windings are closely coupled the resultant
magnetization of the core is of a small order. Since the two wind-
ings are connected “ series aiding ”’ so far as a.-c. currents are con-
cerned, the total inductance is increased. Not only less iron is
necessary but less copper as well.
262 AUDIO AMPLIFIERS
The output resistance of such an amplifier is double that of the
single tube; therefore when worked into a low impedance load an
output transformer must be used to see that maximum undistorted
power is fed to the load. In other words the plate load must be
matched to the plate resistance of the amplifier by means of an
appropriate output transformer.
The output device for the push-pull amplifier may have at least
two forms. Hither it is a straight transformer with two windings
of the proper turns ratio, or it is a center-tapped choke. At first
thought one would judge that the terminals of the loud speaker
would be at high voltages from each other because they are
attached to the plates of the tube, but this is not the fact. It is
true that they are at high potential with respect to earth or to
minus A and so one may get a severe shock if the voltage is high
and if one of the terminals of the speaker is touched by anyone who
is in contact with ground. But the two ends of this choke are at
the same d.-c. voltage and so no d.-c. voltage exists across the
speaker. This situation, of course, prevents any current from
flowing through the speaker. If one desires protection against the
high d.-c. voltage from one speaker terminal to ground, he may
isolate the loud speaker from the plates of the tubes by means of a
condenser, but even here there are large a.-c. voltages developed,
particularly when a percussion instrument in the orchestra to which
one is listening is hit a sharp blow.
The push-pull amplifier, then, is a device for eliminating the
second harmonic distortion which occurs when tubes are worked
too far down on the curved part of their characteristic.
214. General conditions for voltage and power amplification.—
In general, voltage amplification must always take place between
a source of voltage in a low impedance circuit and a receiver of a
voltage which is higher in impedance. The turns ratio of a trans-
former for the greatest voltage amplification is given by N =
V/ Z;/Z,, where Z, is the impedance into which the secondary looks
and Z, is the impedance of the transmitter. If these two impe-
dances are equal the transformer must be a one-to-one ratio affair.
There will be no step-up in voltage.
Where voltage amplification is the goal the greatest am plifica-
POWER NECESSARY FOR GOOD LOUD-SPEAKER OPERATION 263
tion will be attained when working from a very low impedance
device into a very high impedance device, for example a tube with
a low plate resistance working into a tube with a very high grid-
filament resistance. This means that the grid of the following tube
must never be permitted to go positive, for then the input
resistance of this circuit becomes quite low and the amplification
falls and amplitude distortion results.
Where quality of reproduction, that is, a flat frequency charac-
teristic is desired, best results can be attained between two low
impedance devices, for example a tube of low plate resistance
working into a tube of low grid-filament resistance. In other
words high voltage amplification and high fidelity of response are
at odds, especially when it is desired to get both within a few stages
of audio. One may get as much voltage amplification out of a
single tube with a uw of 6 and a transformer with a turns ratio of
12 as he can out of two such tubes and a transformer with a turns
ratio of 2, but in the second case he will get a flat frequency charac-
teristic from 60 to 6000 cycles whereas in the other case he will get
a characteristic that is flat nowhere, and in fact is very sharply
peaked.
When one wants high amplification and high quality he must
get it in several stages, and not all at once.
When one speaks of power amplification the conditions are
not the same as for voltage amplification—in general. Power
amplification may take place only by the aid of a tube which
releases power from a set of local batteries, or from a power line.
The maximum power is transferred when the impedances of the
transmitter and the receiver are equal whether these impedances
are high or low. But in general, power amplification implies the
use of a tube which has a fairly low plate resistance which is worked
into a fairly low impedance circuit or its equivalent secured by
means of a transformer.
215. Power necessary for good loud-speaker operation. —In
1928 about 95 per cent of all commercially manufactured receivers
used the 171 type of tube. The year 1929 saw the introduction of
the 245 type and the still larger 250 type which provides a much
greater margin of safety against overloading. The use of push-
264 AUDIO AMPLIFIERS
pull amplifiers is another step toward protection against overload-
ing and as such is to be encouraged. In the average home push-
pull 245 type tubes provide sufficient power output and a factor
of safety against overloading.
216. Use of tubes in parallel When one cannot use power
tubes of the 245 class, or when one wants more power than a single
tube of any type can deliver, two tubes with grids, filaments, and
plates in parallel may be used. The advantage is that the plate
resistance is halved and that with a given input twice as much
power output can be expected. Two 112’s in parallel on 135 volts
will deliver to a 2500-ohm load 240 milliwatts with a plate current
consumption of 14 ma., which is within the realm of economical
B battery operation.
217. Comparison of push-pull and parallel tubes.—In calcula-
tions involving push-pull and parallel tubes one uses the formulas
developed in the previous chapter for voltage amplification and
power output. The only difference lies in the plate resistance which
is one-half that of a single tube in the parallel case and twice that
of a single tube in push-pull case. Thus if two 112 type tubes are
operated with grids, filaments, and plates in parallel, the plate
resistance will be about 2500 ohms or about the same as that of a
171 tube. The resistance of two such tubes operated in push-pull
will be 10,000 ohms. This means that an output transformer of
the proper turns ratio is necessary when operating the tubes in
push-pull into a low-impedance speaker if the maximum transfer
of energy at some low frequency is desired. Owing to the push-
pull feature of balancing out the second harmonics incurred in
working a high-resistance tube into a low-resistance load,
the output will be relatively free from distortion of this type.
The curve in Fig. 145 shows the effect on the transfer of power
of operating a tube into a load lower in resistance than
itself.
If the input grid voltage is r.m.s., the following formulas give
the power output when the proper turns ratio transformer is used
so that the load resistance is double the resistance of the amplifier.
In these formulas R, is the plate resistance of a single tube, not
of the amplifier.
COMPARISON OF PUSH-PULL AND PARALLEL TUBES 265
2H 2
(1) Single tube power output = ene :
ote
4 p2E2
(2) Parallel tube power output = eae
oiler
4 pH?
(3) Push-pull tube power output = * R —
D
Because of the greater freedom from objectional harmonies
when the push-pull amplifier is overloaded, somewhat greater input
voltages can be tolerated than with a single tube. It must not be
supposed, however, that five or six times the power output of a
single tube can be obtained without serious distortion as has been
stated by some writers. It is probable that the greatest advantages
of this type of amplifier lies in its ability to transmit ordinary
volumes without appreciable distortion, to be operated on alter-
nating current without appreciable hum, and to be used with
plate voltage supply units in which the degree of filtering is not so
great as is necessary with single-tube amplifier stages. For exam-
ple with the push-pull amplifier, no a.-c. currents can get into the
plate voltage supply and it is not necessary to filter the C voltage
equipment so much as with a single tube.
CHAPTER XII
THE DESIGN OF AUDIO AMPLIFIERS
Tue audio amplifier is at least half of the modern radio receiver.
In addition to being necessary to the reception and reproduction of
radio signals, the audio amplifier may be used with phonographs,
talking films, public address systems, etc. The design of audio
amplifying equipment forms a large part of the work of any radio
engineer. This chapter gives some of the theoretical and practical
work that must be understood before one can intelligently design
an amplifier.
218. The transmission unit.—When one compares the voltage
amplification of the power output of any system in which the ear
is likely to play a part—as in an audio amplifier—it is convenient to
express the greater amount of amplification, or power, which one
amplifier gives over another by means of a unit that bears some
relation to the sensitivity of the ear. For example one amplifier
may turn into a loud speaker a power output of 800 milliwatts and
another an output of 1000 milliwatts. How much difference would
this make to the ear? Offhand it seems that a considerably greater
volume would result. But such is not the case. Such a ratio of
one power to another as 1000 to 800 is scarcely discernible to the
ear.
We can state that an amplifier has a voltage gain of 50 and that
under some other condition it has a gain of 60 and imagine that the
latter is easily noted by the average ear. But is it?
A convenient unit of loss or gain is the decibel, abbreviated as
DB, which has taken the place of the TU, an American unit
originated in the telephone industry where nearly all calculations
of power must in one way or another involve the ear. The Bel
was the unit universally adopted in 1928—it is ten decibels—
266
THE TRANSMISSION UNIT 267
and is named in honor of Dr. Alexander Graham Bell, the
inventor of the telephone. The difference in two powers differing
by one DB is just discernible to the ear.
The DB is a logarithmic unit—--that is, each time the amount of
power of a device is doubled—or multiplied by 2—we add 3 DB.
When the power is increased ten-fold—multiplied by 10—we add
10 DB. Here, then, is the second advantage in the DB. We can
add them, instead of multiplying them. For example let us
suppose the voltage amplification of an amplifier is 25 and that it
is to be connected after another similar amplifier. What is the
voltage gain? Evidently, if the second amplifies what the first
gives it, the overall gain will by 25? or 625. Here we must multiply
25 by 25, which is awkward. But if we knew that one amplifier
had a voltage gain corresponding to 25 DB and was to be used
after another of similar characteristics, we would state that the
overall gain was 25 plus 25 or 50 DB. The DB is defined as “ ten
times the common logarithm of the ratio of two powers,” or
Nog = 10 logio (P1/P2) (1)
in which W is the number of DB by which the two powers P; and
Pz» differ. The table below gives some easily remembered values
of DB and their corresponding power or voltage and current ratios:
Approximate
Voltage or
Current Ratio
Approximate
N ;
pune Power Ratio
AL
0
or
2
2.
4
5
8
10
Let us consider an amplifier with an output power of 100 milli-
watts. How much must we increase its power output before the
268 _ THE DESIGN OF AUDIO AMPLIFIERS
ear can just detect the difference? Suppose we double the output.
The ratio is then 200/100 or 2 and the DB corresponding to this
power ratio is 3. The ear can detect one DB difference and using
a table of DB or a slide rule we find that 1 DB corresponds to a
power ratio of 1.25 roughly. Thus the power output to which 100
milliwatts must be increased before the ear can detect the differ-
ence is such that P = P2/100 = 1.25 or 125 milliwatts. Adding
another DB brings the level up to 160 mw. and another unit brings
it to 200 mw., as indicated above.
In this case it would have been foolish to go to great efforts
to effect an output of 115 compared to an output of 100—because
the ear could not tell the difference. In fact the ear can only with
some difficulty tell the difference between the amplifiers differing
by 3 DB—double the power—unless single tones are used and then
only in a quiet room.
219. Voltage and current ratios.—Strictly speaking the DB
should be used only when expressing the ratios of powers. Let us
suppose two amplifiers are feeding current into equal resistances.
The currents are different. How can we express in DB the advan-
tage of the one as a current amplifier? We need only find out the
ratio of the powers as before, and multiply the logarithm of this
ratio by 10. Thus,
P= ak
Pz = Ip? R
DB = 10 log P;/P2 = 10 log i
= 10 log 3
= 20 log 7 (2)
If the resistances are not equal (2) becomes
DB = 20 log 2V Fa _ 20 log Ei/V Ri _ 20 log EV R2
Tov fe Eo/V Re ExV Ri
(3)
VOLTAGE AND CURRENT RATIOS 269
The factor 20 arises from the fact that when one squares a
number the logarithm is doubled. For power ratios, the DB is
10 times the logarithm, for current or voltage ratios the DB is
20 times the logarithm of the ratio.
Voltage or current ratios can be translated into DB only when
the impedances into which the current flows, or across which the
voltage exists, are taken into account. If these impedances are equal
tor both currents or both voltages, they cancel out, one being in
the numerator and one being in the denominator, but in general
they do not cancel out and must be considered.
The DB is always an expression for a ratio. We cannot speak
of an amplifier that has an output of so many DB, but if we assign
some arbitrary level—say 10 milliwatts—and compare all amplifiers
to this amount of power we can say that one has 20 DB or 100 DB
greater output, or less output, or is “ up ” or “ down” 20 or 100
DB. All these DB are expressions for the ratio between these
powers and the “ zero ”’ level power of 10 milliwatts.
Example 1-12. An amplifier has 1 volt applied to its input resistance
of 10,000 ohms. Across its output resistance of 4000 ohms appears a voltage
of 40. What is the power gain in DB? The voltage gain? Would it be
worth while to increase the output voltage from 40 to 50 volts?
Solution.
Powetinplte:! , = 10—* watts.
Power output P, = — = —— = —— = 0.4 watt
sa ls = 4 x 10% = 4000
P; 10-4
Power gain = 10 log 4000 = 36 DB (because the log of 4 is 0.6
and the log of 1000 is 3 and the log of 4000 is 3.6)
BoV Ri
Voltage gain = 36 DB = 20 log ie
t (0)
270 THE DESIGN OF AUDIO AMPLIFIERS
H i = Rs 1.8
ence og — =1,
EV Ro
/R.
Bie = antilog 1.8
E;V Ro
Voltage gain = 63
If EZ, becomes 50 volts,
The gain due to this increased output over Py (above) is
625
in = 10 log —— = 10 log 1.56
gain 08m og
= 2.0 DB (approximately).
And so the gain due to increasing the output from 40 to 50 volts—or from
400 to 625 milliwatts—will be audible to the ear, but the difference is not
worth a great deal of effort to attain it.
The solution of the above example is characteristic of the solu-
tions of all such problems. Given the power ratio it is only neces-
sary to look up the logarithm of this ratio to get the DB gain.
The student must not forget that all numbers between 100 and
1000 have as the first digit of their logarithms the number 2.
Hence all power ratios between 100 and 1000 lie between 20 and
30 DB. Multiplying any power by 10 represents a gain of 10 DB.
Thus of two amplifiers having outputs of 50 and 500 watts, the
latter is said to be 10 DB better than the former. A loss of 10 DB
means that the power in any circuit has been divided by 10. If it
is decreased or increased by 100 times the loss or gain in DB is
20 DB.
Example 2-12. A certain amplifier has a characteristic such that at 100
cycles its amplification in voltage is 8, at 1000 cycles it is 80, and at 6000
cycles, where the amplifier tends to “ sing,” the voltage amplification is 200.
Are these differences appreciable to the ear?
Let us take the amplification at 1000 cycles as a zero level and find out
how much above or below this level the other frequencies are. At 100 cycles
THE USE OF THE DB 271
the voltage ratio is 80/8 or 10. At 6000 cycles the voltage ratio is 200/80 or
2.5. At 100 cycles there is a loss, at 6000 cycles there is a gain. Thus,
80
Loss at 100 cycles = 20 log a 20 log 10 = 20 DB.
8 DB.
200
Gain at 6000 cycles = 20 lo Sate 20 log 2.5
Such a characteristic indicates a poor amplifier. The low notes would be
totally lost and high ones would overload the last tube.
Example 3-12. In a certain circuit there is a loss of 25 DB. What power
ratio corresponds to this loss?
Power ratios of 10 = 10 DB, 100 = 20 DBand 1000 = 30 DB. There-
fore the power ratio of 25 DB lies somewhere between 100 and 1000. The
figure 2 of 25 DB tells us that the loss is somewhere between 100 and 1000
times. The figure 5 of 25 DB is 10 times the logarithm of 3.1 and so 25 DB
corresponds to a power ratio of 310.
The solution of such a problem is as follows:
1
ety 2
25 DB = 10 logio ==
Ps,
Pai eens :
PA = lloye 5 (dividing both sides by 10)
A = antilog 2.5
P,
= antilog 2.0 times antilog 0.5
100 X 3.1 = 310
If the loss were a voltage loss of 25 DB the solution would be:
E,
25 DB = 20 log =
BE,
1925 = loys = (dividing both sides by 20)
2
Ey : : F :
z7 antilog 1.25 = antilog 1.0 times antilog 0.25
2
—10 > aleou—ali(.3
220. The use of the DB.—The transmission unit may be used
to express any ratio of power, voltage, current, mechanical loss or
272 THE DESIGN OF AUDIO AMPLIFIERS
gain, etc. Thus we may say that symphony orchestra has a range
60 DB in power. That is, when it is playing very loudly, fortissimo,
it is 60 DB louder than when playing very softly, pianissimo.
This corresponds to a power range of one million to one. In the
wire circuits which carry the microphone currents from the
symphony hall to the broadcast station, the weakest of the desired
signals must be 40 DB above the noise in the line. The very weak
passages of the orchestra are built up by local amplifiers until the
currents are greater than the noise currents. The limit to the
louder passages is the overloading of the amplifiers either at the
hall or in the broadcasting station. And so the stronger passages
are cut down.
Whenever a circuit suffers a loss in power or voltage or current,
we may express that loss in DB. The frequency characteristic of
an amplifier, or a loud speaker, or of a telephone line may be
expressed in DB by plotting a curve in which zero level is the
amplification or power output at some arbitrarily chosen fre-
quency. Thus if we chose 1000 cycles as a reference frequency, all
other frequencies are either up, down, or flat with respect to the
level at 1000 cycles.
Problem 1-12. What in DB corresponds to a voltage ratio of 100? Power
ratio of 100? What voltage ratio corresponds to 100 DB? What power ratio?
Problem 2-12. A current of 0.006 ampere flows through a resistance of
1000 ohms. A switch reduces this current to 1.0 milliampere. How much is
the current reduced in DB?
Problem 3-12. An amplifier has a normal output of 1 watt. A switch is
provided that its output can be reduced in 5-DB steps. What is the output
in watts when it is reduced by 5, 10, 20, and 25 DB?
Problem 4-12. An amplifier has its power output reduced by 25 per cent.
Is such a reduction in power audible to the ear?
Problem 5-12. A radio receiver has a voltage gain in its radio-frequency
amplifier of 50 DB. Express this in voltage gain, and in power amplification
provided that the same impedance closes the input and output of the amplifier.
Problem 6-12. A radio receiver is so adjusted that a station 20 ke. off
the frequency at which the receiver is tuned is reduced by 40 DB in voltage
below the station that is being listened to. What is the ratio in voltage
between the desired and undesired station?
Problem 7-12. A broadcasting station increases its power from 500 to
5000 watts. What is this in DB? What is the increase if the power is
increased to 50,000 watts?
THE USE OF THE DB 273
Problem 8-12. If an audio amplifier has two stages and each stage has
a gain of 25 DB, what can the gain of the receiver be reduced to when listening
to the 50,000-watt station compared to the 500-watt station provided they are
equidistant? In other words, a given station increases its power from 500 to
50,000 watts. How much audio gain in DB is this worth to the listener?
Problem 9-12. The noise on a certain telephone line is 40 DB down
from the broadcasting signals. What is their power ratio? If the telephone
currents are of the order of milliamperes, what are the noise currents?
Problem 10-12. Phonograph records with single frequencies are made by
the Victor Phonograph Company for use as frequency standards. A certain
record is labeled as being ‘“‘— 2.0 TU ” compared to a certain arbitrary level.
What is the voltage ratio of the record compared to the arbitrary zero level?
Problem 11-12. The maximum power output from a 112 type of tube is
120 milliwatts. The 171 has an output of 700 milliwatts. How much greater
in DB is the 171 power output?
Problem 12-12. A loud speaker is 5 per cent efficient and requires 1.5
watts to give sufficient volume output. If it is made 50 per cent efficient
how much can the power input be reduced to give the same outrut?
Problem 13-12. A tube has a plate resistance of 5000 ohms. Calculate
the power into a load which varies from 1000 to 20,000 ohms and convert the
ratio between the power at maximum to the power at other values of load
resistance in DB. How great can the difference between the load and the
tube resistance be before the ear will note the difference?
Problem 14-12. The sensitivity of a condenser transmitter (high quality
microphone used in better broadcast studios) is 0.35 millivolt per dyne of
force exerted by an air wave impinging on each square centimeter of the
diaphragm. The carbon button microphone has a sensitivity of 5.0 millivolts
per dyne per square centimeter. How much more sensitive is the latter over
the condenser transmitter, expressed in DB? How many stages of trans-
former-coupled audio amplification using 2 : 1 transformers and tubes with a
u of 8 will be required to bring the output of the condenser transmitter up to
the level of the carbon button microphone? A commercial telephone trans-
mitter—such as is used in ordinary telephones—is tuned to the average speech
frequency and therefore amplifies what corresponds to its input about 1000
times. Express in DB its sensitivity compared to the other two microphones.
Problem 15-12. A radio receiver is so adjusted that its output is 8 DB
above an arbitrary level. The maximum power the receiver can put out is
10 DB above this level. If the output power is proportional to the square of
the power of a broadcasting station which is producing the output of 8 DB,
by how much will the output tube of the receiver have to be increased in DB
if the transmitter doubles its power?
Problem 16-12. A radio receiver is tuned to a certain distant station
which gives at the receiver input a voltage of 500 microvolts. A nearby station
on a different frequency produces a voltage of 50,000 microvolts at the same
274 THE DESIGN OF AUDIO AMPLIFIERS
time. How much loss in DB must be put into the receiver at the frequency
of the undesired station to reduce the signals to the same level? How much
additional loss must be put into the receiver to reduce the unwanted station
to 60 DB below the desired signal? The curves in Fig. 166 will be interesting
in connection with this problem. They were published by Lloyd Espenschied
in the Bell System Technical Journal, January, 1927. They show the relative
selectivity of several types of receiver. The “double detection ”’ receiver is a
superheterodyne.
Problem 17-12. It is desired to make a gain control which will affect the
output of a receiver in steps too small to be noted by the ear. An average
100
Decrease in Response D.B.
120
140
10 20 30 40 50 60 70 80 90 100
KC off Resonance @ 900 KC
Fig. 166.—Selectivity of several circuit combinations.
1= Single circuit non-regenerative
2= Couple circuit non-regenerative
3= Single circuit regenerative
4= Tuned Radio Frequency
5= Tuned Radio Frequency
6= Double Detection (Super Heterodyne)
7 = Ideal Characteristic
person will readily note a change of volume of 3 DB in the middle of the audio-
frequency band, but changes of 10 DB at the extreme upper and lower fre-
quencies are not noticed. Suppose the loud speaker has an impedance of
4000 ohms, and that the gain control is to be placed across it—where it should
never go! What shunting resistance will be necessary in the gain control so
that the changes in volume will occur in steps too small to be noticed? The
gain control has a maximum value of 20 DB. ;
Problem 18-12. The ratio of peak power in the voice (accented syllable)
to average may be 200 to 1. Thus if the average power is 10 microwatts, the
peaks may be as high as 2000 microwatts. Express the range in power of the
human voice in DB?
TRANSFORMER WORKING OUT OF HIGH IMPEDANCE 275
221. Design of audio amplifiers.—There have been two schools
of thought regarding the design of radio equipment, especially the
part of the radio set that follows the detector or demodulator,
the audio amplifier. Should each unit be made so that its fre-
quency characteristic and voltage or power gain may be as good
as possible, or should the entire amplifier have the characteristics
desired by permitting some faults in one part of the circuit to be
corrected in another? As long as parts are obtainable so that the
experimenter or the service man can construct his own amplifier,
he must face this problem. But a manufacturer who is not inter-
ested in selling transformers by themselves, but who is interested
in an amplifier which from its input to its output shall have certain
characteristics, need not worry about individual stages. What he
wants is the end product to be as he has planned. It is obviously
unwise to let faults creep into one stage that must be corrected in
‘another if the correction is more difficult than to design the indi-
vidual stages units carefully in the first place, but it is equally
unwise to take great pains to make individual stages perfect when
minor faults in one can be easily corrected in another.
The amplifier designer, then, is interested in the ratio of what
he gets out to what he puts into his amplifier, both as regards
quantity and quality. He must, of course, have an eye on his
individual units as well as on the amplifier as a whole. Let us
begin, then, at the detector or input to the amplifier and work
toward the loud speaker, which we assume is the device into which
the amplifier works.
222. Transformer working out of high impedance.—Some
manufacturers make transformers of different ratios, usually two,
a low ratio to work out of a detector, and a higher ratio to work
out of the first audio tube. Why is this?
A tube has a lower plate resistance the higher the voltage on its
plate and the lower the negative voltage on its grid. A 201—A type
tube operating as an amplifier under normal conditions, that is,
90 volts on the plate and minus 4.5 on the grid, has a resistance of
about 12,000 ohms. A detector tube usually has a much higher
plate resistance because the plate voltage is lower. The normal
detector plate voltage is 45 although many detectors are operated
s
276 THE DESIGN OF AUDIO AMPLIFIERS
with but 22.5 volts on the plate. In addition many detectors, of
the C bias type, have large negative biases put on the grid. The
result is that instead of a resistance of 12,000 ohms out of which
the amplifier must work there is a resistance of from 20,000 to
40,000 ohms.
We have already seen that a large part of the a.-c. voltage in
the plate circuit of a tube will be usefully impressed on the load
resistance if that resistance is several times the tube resistance.
If we couple the detector tube to the amplifier through a resistance,
say of the order of 100,000 ohms, we shall impress a good share of
the detector plate a.-c. voltage on this amplifier—and it will be
more or less independent of frequency. But if we use a trans-
former which is a complicated combination of inductance, resist-
ance, and capacity, the tube and transformer will have a frequency
characteristic which shows little amplification at low frequencies,
high at middle frequencies and little again at high frequencies
unless we are careful to get the correct circuit constants.
A transformer to work out of a high impedance and to give a
good characteristic from 100 to 5000 cycles must have a greater
primary inductance than a transformer which will work out of a
lower impedance. The following are the reasons why such a trans-
former has a low turns ratio. The number of turns that can be
placed on a secondary without increasing the distributed capacity
to a prohibitive figure is limited. This means that the turns ratio
is controlled entirely by the number of turns on the primary.
Increasing the number of turns to get a high inductance decreases
the turns ratio. The next transformer, on the other hand, works
out of an amplifier tube, and hence out of a lower resistance. Its
primary turns can be reduced somewhat and still give a good fre-
quency characteristic, and so the turns ratio will be somewhat
greater. The Amertran DeLuxe transformers have ratios of 3 to 1
and 4 to 1 for the first and second stages. The Sangamo Type A
transformers, on the other hand, are of 3 to 1 ratio and can be
used for either first or second stage. Both sets of transformers
give very good frequency characteristics. Of two transformers,
choose the one with the highest primary inductance for the first
stage. This will be the lower ratio transformer.
RULES FOR THE AMPLIFIER DESIGNER 200
223. Rules for the amplifier designer.—In general the rules the
amplifier designer must follow are these:
1. He must so design the amplifier that no stage can overload
even on the strongest signals that are to be received. This implies
that each tube has its proper C bias and plate voltage.
2. In the plate circuit of each tube must be a sufficient impe-
dance that at low audio frequencies the characteristic is straight
and not curved. This implies that the #,-E, curve for a circuit
may be one thing for 1000 cycles and another for 100 cycles—
which is correct when apparatus is used that has reactance. At
1000 cycles the tube works into one impedance—at 100 cycles into
a much lower one.
3. He will get a better characteristic—although less gain—from
low impedance, low » tubes. This will require more tubes, of
course, to get a predetermined amount of amplification.
4. He will get more amplification—but a poorer characteristic
with high-y, high-impedance tubes.
5. The more tubes in the amplifier, the greater will be the
noise in the output due to “ tube noise.”” One well-known physi-
cist-engineer has stated that the proper place for the loud speaker
is in the detector circuit.
6. The more tubes and the higher the overall gain of the
amplifier the greater will be troubles from instability and from
unwanted pickup from nearby a.-c. magnetic fields.
These are general rules and statements, and it will not pay to
be dogmatic about them. They may work in some instances, and
fail in another.
The maximum voltage amplification that can be secured
from a transformer-coupled stage is » times the turns ratio of the
transformer. If the impedance into which the tube works—the
reactance of the primary if the secondary is open-circuited—is
two times the plate resistance of the tube, 89 per cent of the u of
the tube will be realized. If the impedance of the transformer at
the lowest frequency it is desired to amplify is two times R,, the
amplification at this frequency will be 75 per cent of the maximum
possible, and nearly 100 per cent of the maximum possible will
278 THE DESIGN OF AUDIO AMPLIFIERS
be attained at all frequencies. Then the range in amplification
will be from 89 per cent to 100 per cent at all useful audio fre-
quencies. The difference between these values of amplification will
not be audible to the ear.
224. Comparisons between amplifiers.—The only fair test be-
tween two amplifiers is made by means of a switch which alternately
throws one set-up and then the other to the detector tube or phono-
graph pick-up being used. A test on one amplifier at one time and
a test on another at some different time is no test at all. The music
may be different, the mood of the listener may be different, and
there are too many other variables to give any faith in such a test.
A simple four-pole double-throw switch will throw the amplifiers
to the input and to the loud speaker with but a second’s delay.
The engineer must remember that the ear can scarcely detect
volume differences in which the power ratio is 2 to 1 and that, at
the two extremes of the audio-frequency range, differences in
power of 10 to 1 are none too easily noted. To spend great effort
and much money in making an amplifier flat from 50 to 5000
cycles is not usually warranted by the difference in fidelity noted
by the ear over an amplifier that is down 10 DB at the two ends.
225. Volume control.—The volume of a radio set is seldom
_ controlled by doing something to the audio amplifier. The
volume control is usually before the detector tube, in order that
this tube shall never overload. Such a volume control may go into
the antenna-ground circuit at the very beginning of the signal’s
path through the receiver. Or it may come in one of the radio-
frequency amplifiers or even in the input to the detector. Study
of detection indicates that less distortion will result if this input
to the detector is kept at some constant desired level.
In 1929 it became necessary to develop new means of con-
trolling volume. Many designers not only reduced the voltage
amplification to reduce volume but at the same time reduced the
coupling to the antenna. Some reduced the gain of the audio
system too so that the hum inherent in an a.-c. operated set was
reduced simultaneously with the volume of desired signals.
If the amplifier is used with a phonograph pick-up, some means
of controlling volume may be necessary. Such a control may be
PROPER C BIAS FOR POWER TUBES 279
a variable resistance across the secondary of the first transformer
as shown in Figs. 167 and 168. This has the effect of lowering the
turns ratio of the transformer, and therefore permits only part
of the total available voltage to get to the grid of the amplifier
tube.
The usual output voltage of a phonograph pick-up is of the
same order as that from a detector tube, that is, from 0.1 volt to
perhaps 1.0 volt. If the lower value, some additional amplifica-
tion may be necessary to get the desired volume, and then a means
may be provided whereby the pick-up voltage is impressed on the
grid of the detector tube which is then used as an amplifier, or
another additional transformer or complete stage may be thrown
into the circuit as desired.
rail
Fic. 167.—Shunt resistance volume _ Fig. 168.—Potentiometer gain or
control. volume control.
226. Proper C bias for power tubes.—The proper value of C
bias for an amplifier tube is determined partly by the input volt-
ages to be encountered and partly by the amount of d.-c. power
the plate can dissipate safely. This power is the product of the
plate voltage and the plate current, and the latter is controlled
to some extent by the C bias.
Problem 19-12. A power tube’s plate current is as follows,
Ey E, Ip, ma.
450 0 100
20 55
30 35
40 18
and the maximum safe power that can be dissipated at the plate is 10 watts.
What is the minimum value of grid bias (E,)?
Output Volts R. M. S.
280 THE DESIGN OF AUDIO AMPLIFIERS
227. Loss in output transformer.—With the advent of large
power tubes of the 171, the 210, and the 250 types, the necessity of
using two or more tubes in parallel has ceased, so far as the average
experimenter or set designer is concerned. To get considerable
power output, it is advisable to use the largest tube ordinarily em-
ployed in receivers, namely, the 250 type, instead of using several
tubes in parallel. If still greater power is required—and the 250
will turn out nearly 5 watts of audio-frequency power—several
power amplifier stages may be operated in parallel and connected
to whatever load they are to drive.
60 100 200 500 1000 2000 5000 10000
Ff Cycles
Fic. 169.—Transmission characteristic of output transformer.
A single 250-tube amplifier worked up to its limit would provide
sufficient power for about five loud speakers running at consider-
ably above normal output for average apartment volume level.
For each five apartments, then, another 250-tube amplifier could
be installed and driven from the next-to-the-last stage of audio
in the equipment.
Difference in impedance between the tube and the equipment
into which it works can be taken care of by means of the output
transformer. The curve in Fig. 169 shows the transmission charac-
teristic of a Pacent 1 : 1 output transformer designed to work out
of a 210 type or 5000-ohm tube. The loss in power transmission
occasioned by the transformer is small, and the distortion due to
MANNER OF COUPLING TUBE TO LOAD 281
its own frequency characteristic will not be noted by the most
critical ear.
228. Turns ratio in output transformers.—When there are
large impedance differences between tube and loud speaker, a
transformer must be used.
For example the average dynamic type of loud speaker has an
input impedance of from 5 to 10 ohms whereas the lowest resistance
tube in common use is the 171, 2000-ohms. Hence an output trans-
former is necessary to provide maximum energy transfer from the
tube to the speaker. The turns ratio for such a loud speaker
working out of 5000 ohms would be about 25 : 1.
The turns ratio should be such that the tube always works into
an impedance greater than its own plate resistance, in order to
minimize curvature distortion and
its accompanying second §har-
monics. This means the turns
ratio can be somewhat less than
that necessary to “match” the
tube and speaker at the lowest
frequency to be received.
Impedance—Ohms
The Western Electric 540—A W T cOwI00 ES UORIORN SODORNTIE
speaker, as shown in Fig. 170, f Cycles
has an impedance of about 1500 fy. 170—Impedance of cone type
ohms at, say, 100 cycles, and thus loud speaker.
to couple it to a 2000-ohm tube
requires a step-down ratio of 0.87 for maximum power transfer.
229. Manner of coupling tube to load.—Output devices are
used to
(1) Keep d.-c. current from the loud speaker winding;
(2) Prevent serious loss in plate voltage;
(3) Prevent heating the loud speaker winding by the plate
current of the last tube;
(4) Prevent placing a mechanical bias on the loud speaker
armature or moving element;
(5) Adjust serious impedance differences between tube and
speaker, and therefore to improve fidelity and
increase power;
282 THE DESIGN OF AUDIO AMPLIFIERS
and some output devices
(6) Keep the loud speaker terminals at low d.-c. potentials
and prevent shock or burn.
There are two types of output devices. The transformer
proper having two windings insulated from each other is one type;
another is the choke-condenser device. The choke consists of a
large number of turns of copper wire on an iron core. The con-
denser is an ordinary filter or by-pass condenser preferably of
about 2-4 mfd. capacity. The choke should not be smaller than
12 henrys.
Supply
B
Supply
Fria. 171.—In this output connec- Fig. 172.—Here the loud-speaker
tion, loud-speaker currents return currents must flow through the B
to the filament directly. supply to return to the filament.
The transformer by its very nature keeps d.c. from the loud
speaker. Its primary d.-c. resistance is small and so serious loss
in plate voltage is not experienced.
The choke-condenser also keeps d.c. from the loud speaker
because of the condenser. If the choke is tapped, impedance differ-
ences may be adjusted.
The choke-condenser may be connected into the circuit in two
ways, one of which is much better than the other. In one connec-
tion, Fig. 172, the circuit is the same as the output transformer in
that all of the a.-c. currents in the plate circuit of the last tube
must return to the filament of this tube by going through the plate
voltage supply device, whether batteries or an “ eliminator.” In
the other connection, Fig. 171, the a.-c. plate currents in the loud
speaker return directly to the filament and do not go through the
METHOD OF CONNECTING CHOKE-CONDENSER 283
plate voltage supply device. This is a distinct advantage as will
be seen in Section 230.
The equivalent circuit as far as a.c. is concerned is shown in
Fig. 173. The choke represents a shunt Zz across the tube and
loud speaker, and. to prevent
loss in power it must be large
in impedance compared to
the loud speaker. The con- zx z
denser, in series with the loud
speaker, represents a series
loss Z, and must be small in py. 173.—The equivalent circuit of Figs.
impedance compared to the 171 and 172.
loud speaker impedance. The
loss may be calculated by estimating what power would get into
the loud speaker if the condenser and choke were not present, and
then calculating what power would be transmitted if either or both
of these additional units were present.
Rp Zo
Problem 20-12. A 171 type tube is to feed power into a speaker which at
200 cycles has an impedance of 2000 ohms. The 171 has a plate resist-
ance of 2000 ohms. The condenser is 4 mfd. and the choke is 20 henrys.
Calculate the loss in DB of having the choke present, of having the condenser
present, of having both present. What would be the losses if the choke were
only 10 henrys, and the condenser only 1 mfd.? (Neglect phase in calculations. )
230. Method of connecting choke-condenser.— What difference
does it make which way the loud speaker is connected to the choke-
condenser? Let us consider in Fig. 174 a typical case, a two-stage
transformer-coupled amplifier and a detector getting plate voltages
from a common source; the box labeled ‘‘ B supply.” Let us
suppose an a.-c. voltage of 50 appears across the choke at a fre-
quency of 1000 cycles due to an amplified voltage of this frequency
coming from the detector. This voltage will cause an a.-c. current
of about 0.2 ma. to flow through this inductance, which presents
an impedance of about 250,000 ohms to the 1000-cycle voltage.
To get back to the negative side of the filament of the last tube
this current must flow through the impedance of the B supply,
which for purposes of argument is about 80 ohms (the impedance
of a 2-mfd. condenser to 1000 cycles).
284 THE DESIGN OF AUDIO AMPLIFIERS
This current through 80 ohms impresses 0.016 volt on the
primary of the first audio transformer in series with the plate
resistance of the detector tube. Most of this voltage will appear
across the transformer, and will be amplified to reappear finally
across the output choke. If the transformer windings are poled
correctly and if there are three stages getting plate voltage from
the common supply, this voltage will reappear across the output
choke in phase with the 50 volts already mentioned.
Det Ist A.F. 2nd A.F.
B
Supply
Fic. 174.—The proper output connection.
Let us suppose, now, that the loud speaker is connected directly
across the choke, as in Fig. 175. If this speaker has an impedance
at 1000 cycles of 4000 ohms, the effective load in the plate circuit
of the tube is 4000 ohms shunted by 250,000 ohms, or not far from
4000 ohms, neglecting phase angles. Now the 50 volts a.c. across
this load will produce a current of 12.5 ma. which will flow through
the 2-mfd. condenser and produce 1 volt across it which will be
amplified and return across the load in phase as before. If the
voltage gain of the amplifier from primary of the first transformer
to the voltage across the load is 200, the returning voltages in the
two cases will be 2.1 and 125 volts, respectively. In the second
case the amplifier will ‘ sing ’’ because enough of the voltage across
the output is impressed back again on the input and amplified
VOLTAGE LIMITS ON THE OUTPUT CONDENSER 285
enough to return greater than the original voltage. In the first case
the returning voltage is less than the original voltage, and after a
cycle or two of going through the amplifier the returned voltage
will die out.
The advantage is clearly with the connection which returns the
loud speaker currents directly back to the filament of the final
tube and which keeps them out of the B supply.’ The output trans-
former does not have this advantage. The push-pull amplifier
Det. Ist A.F. 2nd A.F. 4.0 Mfd.
L.S. 4000
oO Ohms
Fic. 175.—Regeneration in audio amplifiers is caused by forcing a.-c. currents
through the impedance of the B supply.
keeps a.-c. currents out of the B supply and is more stable for
that reason.
231. Voltage limits on the output condenser.—In case the loud
speaker is connected across the choke the voltages the condenser
must stand are only the a.-c. voltages existing across the choke
because the d.-c. voltage across the choke is small. On the other
hand, if the loud speaker is connected from the condenser to the
filament as in Fig. 174 the voltages across the condenser are not
only the a.-c. voltages but the steady d.-c. voltage on the plate of
the last tube as well. It must be a better condenser in this case.
If the B voltage is 180, the condenser should be able to stand up
under occasional surges of twice this value. Its rating should be
400 volts. If it blows up, the loud speaker will probably be
286 THE DESIGN OF AUDIO AMPLIFIERS
ruined because the entire plate battery voltage of the last tube
will be impressed on the speaker and the choke in series.
If the condenser in case of Fig. 172 blows up, nothing happens
—the choke will have a lower d.-c. resistance than the loud speaker
and most of the current to the tube will go through it in preference
to the speaker.
232. Regeneration in audio amplifiers—The troubles from
regeneration in audio-frequency amplifiers have been largely over-
looked. The simple case cited in Section 230 where the loud
speaker currents went through the impedance of the B supply
device and were impressed back upon the input to the amplifier is
but one example. The case cited below presents some laboratory
data on just what happens when regeneration exists. A.-c. cur-
rents in an audio amplifier must be returned directly to the fila-
ment of the tube they are generated in, and never permitted to
roam around through the wiring, through the B supply device,
through the C bias resistor, etc. Otherwise, regeneration cannot
help being sometimes helpful, sometimes harmful; since it is never
predictable it is best to prevent it.
Regeneration occurs when any impedance is common to two
amplifier circuits. Currents from one circuit set up a voltage
across this impedance. If this voltage is impressed upon a previous
amplifier circuit, regeneration takes place. Coupling may take
place across a resistance, a condenser, a coil, or across any com-
plex combination of these three components of impedance.
233. A case of regeneration.—A two-stage transformer-coupled
amplifier employing Amertran DeLuxe transformers was built and
proper means taken to prevent a.-c. currents from one circuit caus-
ing trouble in another. The circuit diagram is given in Fig. 176.
The C biases for the two tubes were obtained by the plate current
of the tubes in question flowing through 2000-ohm resistors Ro and
Rs. To prevent any a.-c. voltage which might appear across these
resistors from getting into the grid or input circuit of the tube, the
grid circuits were filtered by a large capacity (1 mfd.) across the
resistor C2 and C4 and two high resistances R3 and Rs in series.
Since no d.-c. current flows in this circuit there is no loss in d.-e.
voltage. The C bias is determined by the size of the resistance and
A CASE OF REGENERATION 287
the plate current flowing. The values of bias are immaterial for
our present discussion.
The plate voltages for the first and second tubes corresponding
to detector and first audio tubes were obtained by reducing the
voltage on the power tube by series resistances R; and Rz4 in this
case from 180 volts to 90 and 45.
A.-c. voltages were fed into the amplifier through a 12,000-ohm
resistor to simulate the resistance of a previous tube. The current
into 4000 ohms was read. The input was constant at all frequencies
and under all test conditions. If the output meter did not change,
10000 Ohms 12000 Ohms _y, 4 Mfd.
Co Py
CS DQz
BS, 25
Beat Frequency SC +e
Oscillator 500 0hms & Os eds
© “<—>\_4 ms
C S
Ss
1.0 Mfd.
Oo aE -———2
— C, 145 ahaa)
50000
hs Ohms
4—B O
Ground +220
Fic. 176.—A high quality, well-filtered audio amplifier.
the characteristic of the amplifier was flat. If it did vary, the
changes were indicative of the gain or loss in amplification at the
frequency then being used. We are not concerned with the amount
of amplification now, we are interested only in the characteristic
obtained. The B supply device was a Majestic B eliminator using
a gaseous rectifier tube.
When all the filtering was in the grid and plate circuits and the
loud speaker was connected so that the a.-c. currents were fed back
to the center of the last tube’s filament and thus were kept out of
the B supply (as in Fig. 176), the characteristic was flat from 100
288 THE DESIGN OF AUDIO AMPLIFIERS
to 8000 cycles, was down only 1 DB at 60 cycles, and was up 8.6 DB
at 10,000 cycles where the transformers were going resonant (A)
Fig. 177.
Removing the filtering in the detector plate circuit so that the
45-volt supply came directly from the eliminator dropped the 60-
cycle response 6.6 DB (B). Removing the filtering in the grid
circuit of the first tube had no effect on the low frequencies and
reduced the peak around 10,000 cycles. Getting the 45-volt supply
from the B supply directly (unfiltered), removing the grid filter
resistors (0.5 megohm)
by-passing the plate re-
sistors, reduced the out-
put at 60 cycles by 6.5
DB; placing the loud
speaker across the out-
put choke and keeping
other conditions the same
iA as the last reduced the
50 100 300500 1000 3000 5000 10000 60-cycle output by 15
Frequency Characteristics of
Amertran Amplifier DB CD. and placing 8
Fic. 177.—Effect of filtering the audio ampli- mfd. across the B supply
fier of Fig. 176. from minus to plus 180
brought up the 60 re-
sponse from the last reading so that it was only 12.8 DB down
from the normal output.
With all of the filtering in the circuit, a resistance of 1000 ohms
in series with the B supply had no effect on the characteristic, prov-
ing that if the amplifier is properly filtered its characteristic is
independent of the B supply impedance. Without such filtering
the inclusion of 1000 ohms would distort the characteristic a great
deal and probably make the amplifier sing.
With no filters at all and the loud speaker directly across the
output choke the response at 60 cycles was down 20 DB (E) By-
passing the C’ bias resistor for the last tube by 1 mfd. brought the
output at 60 cycles up so that it was only 16.6 DB below normal,
showing that some of the a.-c. voltage across this resistor was re-
duced by the 1 mfd. condenser, but that the reactance of the latter
FILTERING IN AUDIO AMPLIFIERS 289
was still appreciable compared with the resistor, 2000 ohms. To
keep all of the a.-c. voltage out of this circuit the capacity would
have to be much greater than 1 mfd.
In another case by-passing the C bias resistor in the first audio
stage brought up the response of a two-stage amplifier over 6 DB
at 6000 cycles. A.-c. voltages across this resistor were fed into the
tube so that they were out of phase with the voltages there due
to the grid input and, of course, detracted from the output at this
frequency. This case was one involving the resonance condition
where the capacity across the secondary resonated with the leakage
inductance of the transformer so that a large a.-c. current went
through the C bias resistor. When this resistor was by-passed its
impedance to 6000 cycles was reduced and so the out-of-phase
voltage set up there and introduced into the circuit was de-
creased, bringing up the output of the amplifier at this
frequency.
Whether or not such resistance should be by-passed depends
upon conditions, and upon amplifiers. Sometimes the character-
istic might be improved—but in general it would not. It is safer
to keep a.-c. currents where they belong and to use good filtering
at all times.
234. Filtering in audio amplifiers.—A filter in an audio ampli-
fier, as indicated in the last section, is designed to keep out of a
certain circuit a.-c. currents of a certain frequency or frequencies.
Let us look at grid circuit of the first tube in Fig. 176. Plate
current flowing through the resistance Rz2 causes a voltage drop.
This drop is utilized as the C bias for the tube. If a.-c. currents
flow through the resistance, they, too, cause a voltage drop, and
since this resistance is now part of the grid circuit the voltages are
impressed on the grid. Since the plate and grid a.-c. voltages are
out of phase, these unwanted voltages getting into the grid circuit
from the plate circuit cause a reduction in the amplification.
If the inductance in the plate circuit is 100 henrys and the fre-
quency is 100;cycles, the current through the 2000 ohm resistance
Rp» will be such that for every volt across the inductance there will
be 0.032 volt across the resistor. This voltage is impressed on the
grid of the tube and if multipled by 8 will reappear in the plate
290 THE DESIGN OF AUDIO AMPLIFIERS
circuit as though coming into the system from the outside via the
transformer. Thus a 25 per cent reduction in output will occur.
This is a loss of 2.0 DB in voltage.
Now this loss may be reduced by by-passing the resistor so that
the impedance offered to 100-cycle currents is smaller and so the
voltage there will be smaller. A 1 mfd. condenser C2 has a
reactance of 1600 ohms at 100 cycles and when placed across this
resistor will reduce the voltage there by a ratio of 2000 to 1250.
Much greater isolation of the grid circuit will take place, however,
if a high resistance R3 is placed in series with the grid and a by-
pass condenser C2 is placed as shown in Fig. 176. Still greater
isolation and freedom from unwanted coupling will occur if the
plate circuit is filtered too, either through a resistance R4 or through
a low-resistance, high-reactance choke and, of course, a condenser
C3. Such filtering prevents a.-c. current from flowing through the
C bias resistor. Grid-circuit filtering reduces the effect of such
a.-c. currents as do get into C bias resistors.
In all such circuits, and indeed in radio-frequency amplifiers,
too, the a.-c. currents in the plate should be returned directly to
the filament of the tube in question. They should not be permitted
to go back through any part of the B supply or even through the
leads to it. The condenser should be a part of the amplifier, and
the choke or resistor may be a part of the B supply, although it is
preferable to have it in the amplifier itself. Then the amplifier is
forever independent of its source of plate voltage.
The purpose of the series impedance in such filter circuits is to
impose a high series loss on any a.-c. voltages or currents that may
try to get into the grid or out of the plate circuit. The purpose of
the condenser is to provide a low loss shunt path for these same
voltages or currents to get to the filament. Any a.c. current
that gets through the resistance or choke must be very greatly
attenuated and on arriving at the grid end of such a series impe-
dance it finds an easy path to the filament which is at ground
potential, and therefore does not affect the plate or grid as the
case may be. In case of plate filtering, a.-v. currents find an easy
path to the filament through the condenser and a high impedance
path to the B supply.
INDIVIDUAL TRANSFORMER CHARACTERISTIC 291
Such a filter is helpful in keeping any hum at 60 or 120 cycles
from getting into the amplifier from the B supply.
235. Individual transformer characteristic—An audio-fre-
quency transformer may be looked at as a simple series circuit
which may go resonant at some one or more audio frequencies.
Such a transformer is never perfect; it has some primary and
secondary resistance, which consumes power and puts losses into
the system, and there is always some magnetic leakage (Section
64) even though this may be reduced to a very great extent by
using high permeability cores, etc. The secondary resistance and
leakage inductance may be transferred to the primary circuit,
theoretically, and for purposes of analysis, by simply multiplying
is Sa be 1:N
_ L1=Primary Leakage Inductance
* Lo=Secondary “ kM
ll
Fig. 178.—Equivalent circuit of transformer.
them by 1/N? where N is the turns ratio of the transformer. The
transformer then looks like Fig. 178 in which are the series resist-
ances R,, rg and ri, and leakage inductances L; and Lz and the
shunt mutual inductance M—which should be very high—shunted
by the capacity C of the tube and the windings, and followed by
a perfect transformer which only serves to give the proper turns
ratio N. If the mutual inductance—coupling between primary
and secondary—is very high, it may be neglected in the following
analysis of why a transformer characteristic goes up at high
frequencies.
If the series inductances, caused by magnetic leakage, and the
capacity C form a series resonant circuit, a voltage input E;; of this
resonant frequency will build up a high output voltage £.. This
resonant peak varies from 4000 to 9000 cycles in transformers
usually used in audio amplifiers. The greater the distributed capac-
ity plus tube capacity across the secondary of the transformer, the
292 THE DESIGN OF AUDIO AMPLIFIERS
lower in frequency is this peak. If the series resistances, due to
primary and secondary resistance and other losses, is low enough
the “‘Q” (Section 125) may behigh enough for the circuit to oscillate
at this resonant frequency, or to sing, as amplifier engineers say.
Since the plate resistance of the tube is part of the series re-
sistance, any decrease in R, through use of a low-resistance tube will
increase the tendency to peak or to sing at the high resonant fre-
quency. Such a rise at the high-frequency end of the audio band
may be seen in Fig. 161.
So much for the high audio frequencies. What happens at low
frequencies? At low frequencies the series leakage inductance is
not of importance because of the low impedance to low frequencies.
The ratio between the tube plate resistance and the mutual reac-
tance of the transformer, which is large, determines how much of
the low-frequency voltage generated in the plate circuit of the tube
is usefully applied to the transformer. If the plate resistance is
high, or the mutual reactance is low—a poor transformer—the
low frequencies will be largely lost in the tube and will not be im-
pressed upon the amplifier. If the capacity across the transformer
resonates with this mutual inductance another peak may occur in
the response characteristic of the tube and transformer. This peak
will be at a low frequency, and in some cases may occur as low as
several hundred cycles. After this resonance occurs, there is a
tendency for the response to fall off, and in some poorly designed
transformers a rapid rise in amplification at say 500 cycles is fol-
lowed by an equally rapid drop beyond 1000 or 2000 cycles. This
tendency to drop off at high frequencies is usually overcome by
the tendency to rise due to series resonance between the capacity
and the leakage inductance. Making the plate resistance of the
tube very high drops out all the low frequencies.
Some tricks can be played with individual transformer circuits
to change the response. For example a resistance in the secondary
circuit, either next to the grid or next to the filament (a C bias
resistor for example) will drop off the high-frequency response.
If next to the filament, as is customary for C bias requirements,
the current through the capacity of the secondary of the trans-
former flows through the resistance and sets up an out-of-phase
COMPARISON OF PUSH-PULL AND SINGLE TUBE 293
voltage there which may cut down the high-frequency response
considerably.
236. Comparison of push-pull and single tube.—Figure 179
shows the result of a laboratory test to determine the amounts of
power from a single 210 type of tube compared to push-pull tubes.
It will be seen that when the grid current begins to flow, the
power output begins to drop off, and that the push-pull tubes de-
Voltage Gain
Voltage Gain
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Fic. 179.—Comparison of single and pushpull tube amplifiers.
livered roughly twice the power that a single tube did. These tubes
were fed through an input transformer, and the effect of grid cur-
rent flowing through the secondary winding was the cause of the
larger part of the distortion appearing in the curves. When the
output power is no longer proportional to the input voltage
squared, distortion results. The output is no longer proportional to
the signals then, and larger inputs do not result in equal amounts of
increase in output power. The push-pull amplifier can tolerate
THE DESIGN OF AUDIO AMPLIFIERS
294
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SCREEN-GRID AUDIO AMPLIFIER 295
somewhat greater input voltages before the distortion reaching the
loud speaker becomes too apparent. In other words the push-pull
arrangement raises the overloading limit somewhat, but two tubes
in push-pull will not deliver four or five times as much undistorted
power as a single tube. The great advantage is quietness and free-
dom from filtering bother.
237. Screen-grid audio amplifier—Screen-grid tubes can be
used in audio amplifiers provided high impedances are supplied for
the tubes to work into. This means that the use of high resist-
ances as coupling units is essential. To determine the relative
advantage of screen-grid tubes compared to high-u tubes, ete., the
circuit shown in Fig. 180 was set up, and the output voltage
measured as a constant input voltage at varying frequencies was
supplied to the amplifier. The result is shown in Fig. 180. The
0.02-megohm and 2.0-mfd. condenser in dotted lines is an “ anti-
motorboating ” circuit to keep the amplifier stable when operating
it from a high-resistance source of plate voltage. It is a filter to
keep the a.-c. currents in their proper paths.
The “ Loftin-White ” circuit is a means of utilizing the high-
voltage gain of a screen-grid tube as an audio amplifier. It is so
arranged that the screen-grid tube looks into the high impedance
input of the next tube, usually a power tube. It is a so-called
“direct coupled ” amplifier, meaning that the condenser in the
ordinary resistance coupled amplifier is done away with, with
consequent gain in fidelity. Rather elaborate precautions are
taken to prevent the amplifier from being unstable, rather high
voltages are necessary, but a high gain, flat characteristic amplifier
requiring but small expense and space results.
CHAPTER XIII
RADIO-FREQUENCY AMPLIFIERS
TuerRE are three basic qualities which must be weighed not
only by the designer of a radio receiver but by the purchaser and
ultimate user as well. These are the sensitivity, the selectivity,
and the fidelity of the receiver. The sensitivity of a receiver is an
indication of the overall amplification from antenna-ground bind-
ing posts to loud speaker. A receiver that is very sensitive requires
but a small input voltage to deliver considerable output power.
The selectivity of a receiver is an indication of its ability to dis-
criminate between wanted and unwanted signals. An infinitely
selective receiver would be one that would respond only to a given
station and not at all to another no matter how powerful this un-
desired signal was, nor how close in frequency it was to the desired
signal. That there is no such receiver goes without saying. The
fidelity of a receiver tells how well it reproduces what takes place
in the broadcasting studio. If a certain voltage at 100 cycles is
set up in the broadcasting studio, then an equal amount of 100 cycles
should come out of the receiver, no more and no less. The same
should be true at 1000 and at 10,000 cycles. In other words a
receiver that delivers a perfectly faithful signal is one which has a
perfectly flat audio-frequency response curve from antenna to
loud speaker input. It has a high degree of fidelity.
A receiver that is perfectly selective, infinitely sensitive, and
delivers a perfectly faithful signal is impossible to obtain. A
receiver which is selective and sensitive enough for practical pur-
poses, and the frequency curve of which is so good that the ear would
not detect its infidelity is not difficult to design, to build, or to
operate.
The sensitivity and selectivity are controlled almost entirely by
296
FIELD STRENGTH 297
nue radio-frequency amplifier, which also has something to do with
the fidelity of a receiver although this should really be a function
only of the audio amplifier.
238. Purpose of radio-frequency amplification The radio-
frequency amplifier serves one primary purpose, which is to obtain
sufficient amplification to make up the enormous loss in power
between the transmitting and the receiving stations. It may also
be useful in obtaining selectivity, although it is not necessary that
selectivity and sensitivity be interrelated.
In the majority of receivers in common use, sensitivity and
selectivity are gained at the same time, and with the same appa-
ratus, and a set that is very sensitive is usually very selective. The
fact that the fidelity of the receiver, which is usually considered
only as a function of its audio amplifier, may suffer considerably
at the hands of the radio-frequency designer and is seldom indepen-
dent of selectivity and sensitivity, is a fact that is generally over-
looked.
239. Field strength.—The voltage that is set up across a receiv-
ing antenna is called the field strength of the transmitter at that
particular point on the earth’s surface. It is usually of the order
of millivolts, and because a higher antenna will pick up a greater
signal—that is, the voltage across it will be greater—it has become
standard practice to rate field strength as so many microvolts or
millivolts per meter. Thus an antenna that has an effective height
of one meter and has four microvolts across it is situated in a field
strength of four microvolts per meter. The effective height of the
antenna is somewhat less than its actual physical height above
ground, and in most receiver measurements is assumed as four
meters (13 feet). An antenna that has an effective height of 4
meters and a voltage of 10 microvolts across it is immersed in an
electric field due to some transmitting station whose strength is
2.5 microvolts per meter.
The greater the field strength at a given point the more volume
one can get out of a receiver with a fixed amount of amplification.
Similarly the greater the field strength the less receiver amplifica-
tion is necessary to give out a certain amount of power.
Dr. Alfred N. Goldsmith, in the Proceedings of the I.R.E.,
298 RADIO-FREQUENCY AMPLIFIERS
October, 1926, has given the following tables which are self-
explanatory:
TABLE I
Signal Field Strength Nature of Service
0.1 millivolt per meter poor service
1.0 millivolt per meter fair service
10.0 millivolts per meter very good service
100.0 millivolts per meter excellent service
1000.0 millivolts per meter extremely strong signals
TABLE II
Antenna power Service Range
5 watts 1 mile
50 watts 3 miles
500 watts 10 miles
5,000 watts 30 miles
50,000 watts 100 miles
Now quoting Lloyd Espenschied in the Bell System Technical
Journal, January, 1927, ‘‘ Fields between 5 and 10 millivolts per
meter represent a very desirable operating level, one which is
ordinarily free from interference and which may be expected to
give reliable year round reception, except for occasional interference
from nearby thunder storms.
“From 0.1 to 1.0 millivolt per meter, the results may be said
to run from good to fair and even poor at times. Below 0.1 milli-
volt per meter reception becomes distinctly unreliable and is gen-
erally poor in summer. Fields as low as 0.1 millivolt per meter
appear to be practically out of the picture as far as reliable high
quality entertainment is concerned.”
These figures of Goldsmith and Espenschied give us a good
idea of what may be expected from stations of certain power at
certain distances from the receiver. From these and other data we
may assume that a 5000-watt station may be expected to deliver
a field of about 10 millivolts per meter at distances up to 20 miles
and 1.0 millivolt per meter not over 50 miles. According to the
Bureau of Standards paper, ‘‘ Progress of Radio Measurements,”
April, 1924, ‘‘ When WEAF was transmitting with 3 kw. in the
ADVANTAGE OF HIGH POWER AT TRANSMITTING STATION 299
antenna its field strength at 10 miles was 32 millivolts per meter.
When KDKA had a nominal power of 10 kw. its field at 10 miles
was 43 millivolts per meter.”
The purpose of the transmitting station is to provide a good
lusty signal that will override static and other disturbances; the
purpose of the radio-frequency amplifier is to provide the listener
with good loud signals from the field strengths which the stations
produce.
240. Advantage of high power at transmitting station.—What-
ever voltage exists across the antenna, whether noise or desired
signal, is amplified by the radio-frequency amplifier; there is there-
fore a distinct advantage, so far as the receiver is concerned, in
using large amounts of power at the transmitting station. The
greater the ratio of signal to noise the better will reception be. No
matter how great the voltage gain of a radio-frequency amplifier,
it cannot bring a weak signal out of the noise and give satisfactory
reception. The signal must always be about 40 DB above the
noise level in order to provide an entertainment free from a noisy
background that is apparent on weak musical passages. When-
ever the noise comes up, as on a warm summer day, and the trans-
mitter station power remains constant, reception suffers, and it
suffers in a very rapid manner the farther the receiver is removed
from the station. The noise about a given receiver is more or less
constant under a given set of conditions, whereas the field strength
due a transmitter decreases to one-fourth every time the distance
is doubled.
If a receiver is situated in a quiet locality, where the noise level
made up of stray voltages from street cars, elevators, arc lamps,
power leakages from high tension wires to trees, sputtering flat-
irons, X-ray machines, etc., is weak, the greater the amount of
amplification in the r.-f. amplifier, the greater the distance away a
transmitter of a given power can be and still provide an adequate
loud speaker signal; and, of course, with this amount of amplifica-
tion the weaker a station can be at a given distance to provide this
loud speaker output. The purpose of the radio-frequency amplifier
is the same as of a telescope; it is to decrease the effective distance
between the transmitter and the receiver. Unlike the telescope, it
300 RADIO-FREQUENCY AMPLIFIERS
cannot be aimed at a particular station but must pick up all the
r.-f. voltages not only between it and the desired station but in
other directions as well. If the power of the transmitter is increased
by 10 DB (ten times the power) the r.-f. amplifier can be made
10 DB less sensitive and thus unwanted signals are automatically
reduced 10 DB compared to the desired signal.
241. The task of the radio-frequency amplifier.—The rf.
amplifier employed in a broadcast frequency receiver may differ
decidedly from that used in a receiver serving other purposes or
tuned to other frequencies. For example it is possible to make a
much more efficient amplifier if it is to work at one frequency
instead of at any one of many. A broadcast receiver, for example,
must be capable of amplifying at any frequency between 500 and
1500 ke. It must be easily changed from one frequency to another,
and its amplification at all frequencies within this band should be
uniform. If it selects and amplifies too, its task is much more
difficult to perform, as we shall see.
The energy thrust upon the ether from a given broadcasting
station is a complex bit of wave motion. If the microphone is idle,
what comes from the antenna may be considered as a very narrow
band, at say 600 ke., called the carrier wave. If a single tone, say
1000 cycles, is put into the microphone, the antenna current has
frequencies of not only 600 ke. in it but 599 and 601 as well, and
when music is broadcast the frequencies in the antenna may be
varying between zero and 5000 cycles above and below the carrier
from instant to instant. These frequencies on either side of the
carrier are called the side bands. The characteristics of the
transmitter must be such that each of these audio frequencies is
given equal power compared to the others. The resonance curve
of the antenna system of the transmitter, then, must not be sharp
but must be rather flat or dull as shown in Fig. 181. It must have
a rather flat top from 5 ke. below to 5 ke. above its carrier fre-
quency.
If audio frequencies up to 5000 cycles are transmitted, each
station requires a channel 10 ke. wide for its transmission, and if
there are 1000 ke. available there are 100 channels or places for
100 simultaneous transmissions.
THREE TYPES OF RADIO-FREQUENCY RECEIVING SYSTEMS 301
At the listening station, the receiver must be able to pick out
any one of these stations, and to receive it without being bothered
by others on other channels. This means that a receiver with a
good degree of selectivity is one which will receive, transmit, and
amplify signals on the band from 595 to 605 ke. but not recognize
a signal in the adjacent channels, that is, on the channel extending
from 585 to 595 ke. and the channel extending from 605 to 615 ke.
In other words, to cope with
conditions in the broadcasting 10 KC
band a receiver should have
““ten kilocycle selectivity.”
242. The ideal response
curve of a receiver.—To carry
out this double purpose of the pater
r.-f. amplifier the response
curve should be a_ square-
topped steep-side curve as in
Fig. 181, like the transmitter wthlble.:
curve. This is very difficult Canes
if HOw impossible to attain. In Fic. 181.—A flat-topped steep-sided re-
practice, the response curve is sponse curve is the ideal. This curve
either so broad that stations approaches it.
on the adjacent band, or even
two or three channels away, will be audible during weaker passages
of the desired signals, or so selective that the high audio tones
are lost, due to what is called ‘‘ side band cutting.” This means
simply that the radio-frequency waves corresponding to 600 kc.
plus or minus 5000 cycles are cut off in the r.-f. amplifier.
A receiver which has such a sharp curve that the higher audio
tones are discriminated against cannot possibly deliver a high
quality loud speaker output even though a flat audio amplifier is
used. A receiver that has a flat-top curve usually has too gently
sloping sides so it is subject to interference from unwanted signals.
243. Three types of radio-frequency receiving systems.—The
majority of receivers for use on broadcast frequencies amplify and
select at the same time. They use tuned circuits, and so the
receivers are called ‘tuned radio-frequency sets” or simply,
302 RADIO-FREQUENCY AMPLIFIERS
“tr.” Because of the steepness of a resonance curve, the overall
response of several circuits to frequencies off resonance is dimin-
ished, and is a logarithmic function. That is, if two amplifiers
deliver ten times as much voltage at resonance as they do at some
other frequency the total discrimination in favor of a desired signal
is 100 times in a two-stage amplifier, or 10” if there are N stages.
Another type of receiver, the super-heterodyne or double detec-
tor, changes the frequency of the incoming signal to a lower fre-
quency and then amplifies at this frequency. In the frequency-
changing process a certain amount of selectivity is developed.
In still another type of receiver the selectivity and the amplifi-
cation are developed in different circuits. One can either amplify
and then select or select and then amplify. A series of tuned cir-
cuits is set up which have more or less flat-top and steep-side
response curves, which can be tuned so that a band of certain
width is cut out of the broadcast frequency spectrum at any desired
point. The signals that are passed by this ‘“‘ band selector ”’ are
then built up in voltage by an amplifier that has little or no dis-
crimination, that is, it amplifies all signals, no matter of what
frequency the same amount, depending upon the selector to pick
out the desired signals. Such a selector is called a “‘ band selector”
or band pass tuner because it selects or passes a band of fre-
quencies.
Problem 1-13. The response curve of a single tuned circuit in a radio-fre-
quency amplifier is given in Fig. 182. Remembering that the amplification
of two stages is the square of one stage, calculate and plot the result of using
two such stages. Then calculate and plot the response curve in percentages,
using the response at resonance at 100 per cent. Finally, plot this curve in
DB using the response at resonance as 0 DB; and, remembering that a station
10 ke. off resonance must be about minus 40 DB in voltage if it is not to be
too loud during weak passages of the resonance signal, determine if two stages
of such amplification and selection are sufficient. If not, would another stage
be sufficient? How many DB does a single stage “put down” a signal
) 20 ke. off resonance; how many DB down is it after passing through two
stages; how many after going through three stages?
Problem 2-13. Suppose a station increases its power from 500 watts to
5000 watts and thereby increases its “ service range ”’ (radius) from 10 miles to
30 miles. Suppose the density of population in the area covered by the
station’s signals is 140 people per square mile. Calculate how many more people
RADIO-FREQUENCY AMPLIFIERS IN GENERAL 303
can now hear the station and the saving to the community if each listener
would have to pay $1.00 each to increase the r.-f. gain of his receiver to get
the station properly if it had not increased its power. If it costs the station
$35,000 to make this change in power, has it been economical from the stand-
point of the community?
60 565 570 575 580 585
KC
Fic. 182.—Response curve and diagram of connections of single-screen grid
stage.
244. Radio-frequency amplifiers in general.—In amplifiers
which are to operate at frequencies far above the audio tones with
which Chapters XI and XII dealt, we have, in addition to the
problems encountered there, several new ones. The difficulty in
maintaining amplification with ordinary tubes and circuits at high
audio frequencies was mentioned, and the reason—the stray
590
304 RADIO-FREQUENCY AMPLIFIERS
capacities and those due to the tube input circuit—was discussed.
These stray capacities in a radio-frequency circuit become much
more important, not only for their shunting effect—which is severe
at frequencies of the order of one million eycles—but because of
other interesting phenomena which will be discussed at this point.
245. Effect of tube input capacity——In general a voltage
amplifier must be one which works from a low impedance into a
high impedance load. The problem in all tube amplifying circuits
in which high-voltage amplification—at the expense of low-power
amplification if necessary— is desired, is to get a high impedance
for the tube to work into. Suppose this load impedance is a resistor
shunted by the succeeding tube grid-filament path. The effect
of the input capacity of this tube upon the load impedance was
discussed in Section 198. If the frequency at which the amplifier
is to work is multiplied by 100, the shunting effect of the condenser
becomes 100 times as great.
Problem 3-13. A tube with a plate resistance of 12,000 ohms and a uz of 8
has a resistor in its plate circuit of 50,000 ohms. The following tube is a
UX-—240 with an effective u of 20 and a grid-filament capacity of 4.0 mmfd., a
plate-grid capacity of 8.8 mmfd., and a plate-filament capacity of 1.5 mmfd.
Other capacities across its ouput circuit bring up the total plate-filament
capacity to 6.0 mmfd. What will be the impedance in the plate circuit of the
first tube and what will be the voltage gain at 10, 100, 1000 kc.? (Section 178.)
Now it is generally assumed that if an amplifier tube has a
sufficiently high grid bias that no grid current flows, the input
impedance of the tube is infinitely high and acts as a pure capacity
which is given by the expression
Co = Cop + Cop (G+ 1),
the effective amplification;
where G
Cop
C,; = the grid-filament capacity.
l|
the plate-grid capacity;
|
If this were true at radio frequencies, as is usually true at audio
frequencies, we could find a way to get around the shunting effect
of the input capacity. But at high frequencies this is not exactly
true. The fact that the input impedance is not a pure capacity
TUNED RADIO-FREQUENCY AMPLIFIERS 305
hecomes important. There is resistance as well as capacity in this
input circuit, and this resistance may vary from a high positive
resistance which absorbs power from the previous tube’s output
circuit, to a negative resistance which delivers power to the
previous circuit. |
246. Tuned radio-frequency amplifiers——The serious shunting
effects of the input capacity of the circuit into which a tube works
and which make it impossible to build resistance-coupled amplifiers
to work at frequencies of the order of 1000 ke. can be avoided by
the simple expedient of using this capacity to tune an inductance,
or, stated in another way, by balancing out the capacity reactance
by means of an inductance. For example in Fig. 183 the effect of
the capacity C, is to so re-
duce the impedance in the 9 ww
plate circuit that no ampli-
fication can result. If,
however, we place an in-
ductance across this con-
denser of such a_ value
that at the desired fre- 0——
quency the coil and con- Fia.183.—Balancing out capacity reactance
denser form an anti-reson- by means of shunt inductance (ZL).
ant circuit of very high
impedance, the load into which the tube works will be not
much less than R and high enough to permit some amplifica-
tion. Since fairly low resistance coils are easily obtainable, and
since the effective resistance of such a tuned circuit is L?w*/r or
L/Cor, a value that may be considerably beyond that of the
resistor, we may as well do away with the resistor and use merely
the coil and condenser. For purposes of selectivity we may place
a variable condenser across C, and so tune the coil over the entire
broadcasting—or other—band.
Problem 4-13. Suppose an inductance of 200 microhenrys 1s in the
plate circuit of the tube in Problem 3. Its resistance is negligible in com-
parison to its reactance. Calculate the impedance in the circuit at 1000 ke.
and at 1010 ke. and hence the voltage gain. Then assume that the coil has
a, resistance (r) of 10 ohms, and when tuned by means of a condenser to
306 RADIO-FREQUENCY AMPLIFIERS
1000 ke. calculate the impedance in the circuit at 1000 ke. and at 1010 ke.
Calculate the voltage gain of tube and load, and the relative advantage of the
tuned circuit in selectivity over the untuned inductance.
The advantages of such a circuit are: first, the effect of the
capacity Co is eliminated; second, the effective resistance in the
plate circuit of the tube may be made as high as we desire by
making the inductance large and its resistance low (effective values
of 100,000 ohms are not difficult to attain); and third, where a
resistance amplifier would be absolutely non-selective, this tuned
amplifier can be made very selective.
247. Effect of negative input resistance.—Now, such a tuned
radio-frequency amplifier works out very nicely theoretically, and
practically comes quite close to the final solution, except for one
thing. This is the changing input resistance of the tube.
The input impedance of a tube is not a pure capacity. It is
a capacity and a resistance. The value of this resistance may be
high and positive if the load in the plate circuit of this tube is
negative in sign—a capacity load, or negative if the load is induc-
tive and of sufficient value. For example the curves in Fig. 184
(taken from Bureau of Standards Circular No. 351, Effect of Loadon
Input Impedance of Tubes, by J. M. Miller) show the input resist-
ance of a typical tube at various values of inductance and resist-
ance in the plate circuit. In other words, the minute we put an
inductance in the plate circuit of a radio-frequency amplifier, we
have done something to the grid circuit of that tube. If (1) the
inductance is high enough, we have decreased the input resistance
of the tube to a very low value; if (2) the inductance is increased
still more, the input resistance becomes negative.
Up to the point when the input resistance becomes negative, it
takes power from the circuit to which it is attached, because any.
r.-f. currents flowing through this resistance must suffer an J/?R
loss. But the minute the resistance becomes negative, power is fed
into the circuit to which the tube is attached. Now, feeding power
into that circuit has the same effect as decreasing the resistance of
that circuit by some other means, and when sufficient power is fed
into the circuit that all of its resistance has been reduced to zero,
radio-frequency currents flow although the input is removed from
EFFECT OF NEGATIVE INPUT RESISTANCE 307
its original driving source, and the circuit is said to oscillate.
Direct-current power from the batteries is used up in producing and
maintaining radio-frequency power in the coils and condensers.
The circuit is useless as an amplifier.
We come then to this impasse, that, to get any amplification out
of a tube at radio frequencies, we must put an inductance in its
plate circuit and tune that inductance with a condenser. Putting
Inductance in Plate Circuit Millihenries
2
14
n
£
=
=)
3s |__| __
= f=318000
”
BLA [+=6 Fax
“4 Rp =20000
a Cop = Cop =Cy-
3-4)
-6
\ Ace
Ie NS) oka ie Jind
NA eee
Fig. 184.—Dependence of input resistance upon inductive load.
the inductance in the plate circuit reduces the input resistance of
the tube, and when the inductance is tuned to resonance, the cir-
cuit is in a highly critical condition. Any increase in inductance, or
decrease in capacity—which amounts to the same thing—is liable
to make the whole system oscillate. Any inductance in the plate
circuit reduces the resistance in the grid circuit and results in a
reduction in the effective resistance losses in the circuit out of
which it works. So any resonant circuit that may be present—an
308 RADIO-FREQUENCY AMPLIFIERS
antenna for instance—becomes very sharply tuned, and the higher
audio frequencies are completely lost. The circuit is said to be
regenerating. Regeneration is merely a minor case of oscillation.
For the moment let us forget this trouble from regeneration and
oscillation with the remark that it is caused by the inter-electrode .
capacity of the tube, and is of sufficient magnitude to have pre-
vented efficient r.-f. amplifiers from becoming generally used for
a long time after the need for such equipment was felt. Let us
look into the tuned radio-frequency amplifier, Just as though we
never heard of oscillation or regeneration, and see what engineer-
ing we can bring to bear on the problem of getting the most ampli-
fication and the best fidelity with the apparatus at hand.
248. Engineering the tuned radio-frequency amplifier. The
inductance and capacity required for resonance at any point in
the broadcast frequency band (550 to 1500 ke.) may be calculated
easily. These values are more or less fixed by the sizes of con-
densers generally available. Let us suppose their values are such
that with the resistance of the coil taken into consideration they
produce an effective resistance at resonance L/Cr (an anti-resonant
circuit) of 100,000 ohms. In the plate circuit of a 12,000-ohm tube
this load should give considerable amplification except for the fact
that bridging a resistance of 12,000 ohms across such a coil-con-
denser combination would have the same result as adding consider-
able resistance in series with the coil, and therefore the effective
resistance into which the tube works will decrease at an alarming
rate.
For example, if the effective resistance of a coil condenser com-
bination is 100,000 ohms and is shunted by a 12,000-ohm tube,
the equivalent impedance is now reduced to 10,700 ohms which
represents the same change produced in the selectivity of the anti-
resonant circuit as if its resistance had been increased by some-
thing over nine times. Of course this would give a very poor degree
of selectivity, and the voltage amplification would not be as high
as desired.
If, however, we use a transformer with a step-up ratio from the
plate of one tube to the next grid circuit, the plate resistance is
stepped up by the turns ratio squared and is then placed upon the
ENGINEERING THE TUNED RADIO-FREQUENCY AMPLIFIER 309
tuned circuit. If, for example, the turns ratio between secondary
and primary is 4, the effective resistance placed across the tuned
circuit would be 12,000 x 4? = 193,000 and so the resistance of
this tuned circuit would be increased only by a ratio of 10 to 6.1.
Furthermore, there is a voltage step-up in the transformer, there-
fore some voltage gain may be secured by its use. Of course this
transformer need have only one winding, the input coil to the fol-
lowing tube being tapped as in Fig. 185 for the plate inductance of
the previous tube, an auto-trans-
former, in fact.
In this case the transformer
may be looked at as a kind of
selectivity adjuster, since one
can adjust the selectivity by
its use.
Let us look at the problem in
another way. A transformer with
tuned secondary will give maxi- Fic. 185.—-Proper selection of the
mum power in the secondary load tap will provide maximum voltage
and maximum voltage across the pmpinea en,
secondary when the _ resistance
across secondary and primary are related by the proper turns ratio.
If, then, the load across the secondary is the effective resistance of
the secondary when tuned to resonance, and the resistance across
the primary is the plate resistance R, of the preceding tube, the
proper turns ratio can be found by the usual means, namely,
Z: tlhe Sets
ee en CF nt
whence
Lw
N= )
Vari,
and when this turns ratio is used the voltage gain of tube and
transformer is
310 RADIO-FREQUENCY AMPLIFIERS
This expression involving the turns ratio is not the number
of turns on the secondary divided by the number of turns on the
primary because in an air core transformer of this type it is not
possible to obtain 100 per cent coupling, but for the moment Jet
us not bother about this problem. We shall return to it later.
The actual turns ratio is somewhat greater than N, and can be
determined by experiment.
249. Gain due the tube and gain due the coil.—There are two
parts to the above expression for the voltage gain of tube and trans-
former. Part is due the tube and part is due the transformer.
That is, we may divide up this expression into two parts.,
mM Lw
2 » omy
VR, Vr
in which the first part shows that the gain is proportional to one-
half the » of the tube divided by the square root of its plate resist-
ance and the second part shows that the gain is proportional to the
inductive reactance of the coil divided by the square root of its high-
frequency resistance. From such an expression we may learn
several things.
In the first place G is the maximum possible voltage gain that
can be obtained from a given tube working into a given resistance
and this is obtained only when the turns ratio N of the transformer
ea |
VR,
is a constant and is related to the mutual conductance although
once we have determined its value we can multiply it into the
corresponding values for the voltage gain due the transformer
and we then have the maximum voltage gain. The figure of merit
for the transformer (Lw/ Vr) is not constant over the broadcast
frequency band as the curve in Fig. 186 shows, and from it we
learn at once that the overall gain, G, of tube and transformer
will not be equal at all frequencies unless the overall gain is reduced.
This is because the maximum gain is being secured at the low
frequencies, and there is but one way to flatten the curve, that is,
to decrease the high-frequency gain.
is properly adjusted. For a certain tube, the expression
GAIN DUE THE TUBE AND GAIN DUE THE COIL 311
We are now in a position to perform a very interesting experi-
ment, one that requires neither laboratory nor apparatus. We
need only a pencil, some paper, a slide rule, some values of the
factors that enter into the expression for the maximum voltage
amplification, G.
A eR
20 | 200 — = a
Pars <
, /
proccess rt
: nae mas PS 450
10 p >
CPT ALN +
C7
feet
500 600 700 800 900 1000 1200 1400
Frequency KC
L : ;
Fie. 186.—How resistance and figure of merit “Q” (:*) of a coil vary with
r
/ frequency.
Experiment 1-13. A coil has an inductance of 168 microhenrys, and a
high-frequency resistance, r, of 4 ohms at 700 ke., 6 ohms at 1100 ke., and
9 ohms at 1500 ke. Plot this resistance against frequency. Calculate the
(23)
figure of merit iA) for the coil and plot against frequency. Assume it to
Vr
be used out of a tube whose plate resistance is 12,000 ohms and whose wis 8.
Calculate the proper turns ratio for maximum voltage amplification at the
above frequencies and what the amplification is, that 1s, calculate and plot
Lw
V Rye
turns ratio is used at each frequency.
and G =i, XN against frequency, assuming that the proper
N=
312 RADIO-FREQUENCY AMPLIFIERS
Then calculate N and the voltage amplification at 1000 ke. of such a coil
and a 199 tube with a plate resistance of 16,000 ohms and yu of 6.6, and then
when a 240 type tube is used. It has a plate resistance of 150,000 ohms and
a pw of 30.
Problem 5-13. The effective resistance of a coil-condenser combination
is 100,000 ohms when the resistance of the coil is 10 ohms. That is, L/C r=
100,000 ohms, where r is the series resistance. Calculate what the_ equiv-
alent resistance of such a circuit is when shunted by 10,000 ohms. 1) i wo
resistances in parallel have a resistance equivalent to their product divided by
their sum.) Then assuming that this value is the effective resistance of another
tuned circuit having the same L and C but a es series resistance, cal-
culate what the value of r is. |!
Problem 6-13. A tube of 5000 ohms (UX-1 2A) is to be used as an r.-f.
amplifier and to be worked into a tuned circuit whose effective resistance is
120,000 ohms. What is the proper turns ratio for maximum voltage amplifica-
tion and what is the amplification? Remembering that a resistance ucross a
primary of a transformer is equivalent to another resistance across the secon-
dary stepped up by the square of the turns ratio of this transformer, what, is
the equivalent resistance that is placed across the tuned circuit by this trans-
former? What happens to the effective resistance L/C r of this circuit?
Problem 7-13. A voltage gain of 10 is desired. The plate resistance of
the tube is 12,000 ohms, its » is 15, the inductance of the coil is 186 micro-
henrys, and the capacity is 0.0001 mfd. What is the maximum resistance
the coil can have? Es =34
250. Effect of coupling.—The foregoing argument on maximum
possible amplification of tube and its accompanying transformer
has been based on the assumption that the coupling between
primary and secondary was adjusted to the proper value at each
frequency. Actually, such is seldom or never the case, and even
if it were it is doubtful if receiver designers would so choose the
coupling that maximum amplification would result owing to the
decrease in selectivity of the circuits under these conditions.
In general the voltage amplification of a tube and transformer
when the secondary is tuned may be expressed as
w2M Le
G =. _,
IRR, + wt?
which involves not only the resistance of the apparatus (R, and
R,) on the two sides of the transformer but the coupling M in the
transformer itself. If we go into the laboratory and measure the
EFFECT OF COUPLING 313
voltage across the secondary with a constant input voltage to the
tube but with various degrees of coupling we shall get a curve
similar to that in Fig. 187 (A. I. E. E. Journal, May, 1928,
R. 8. Glasgow, Tuned Radio-Frequency Amplifiers), which is the
experimental proof of our statement that the coupling must be
adjusted for maximum voltage transfer.
Voltage Amp. G
20 40 60 80 100
Mutual Inductance in ph
Fic. 187.—Relation between coupling and amplification.
251. Effect on secondary resistance, of close coupling.—The
increase in resistance—and consequent decrease of selectivity—of
the secondary circuit may be calculated from the expression
_ Mx?
ig
r
which states that the resistance introduced across the secondary by
the transformer is equal to the square of the mutual reactance
divided by the resistance in the primary, in this case the plate
resistance of the tube. This resistance introduced across the secon-
dary increases as the frequency and the mutual inductance squared
and is inversely proportional to the plate resistance of the tube
out of which the transformer works. This means simply that the
greater the impedance of this tube, the less will the selectivity be
decreased with a given turns ratio, and the reason for such an
inverse function is simply that a larger resistance shunted across
314 RADIO-FREQUENCY AMPLIFIERS
the tuned circuit will introduce less resistance in series with that
circuit than will a smaller resistance across the circuit. No matter
what the plate resistance, if M is adjusted for maximum possible
voltage amplification the selectivity of the secondary circuit will
be cut in half, because this selectivity is a function of the series
resistance of the secondary which is doubled when the turns ratio
is such that G is a maximum. Now the coupling between primary
and secondary is controlled not only by the physical proximity of
the two windings, but upon the number of turns in the primary, if
the secondary turns are held constant. Therefore, decreasing the
primary turns will increase the selectivity and decrease the amount
of voltage amplification. Mathematical analysis of the problem
will show that decreasing the turns in the primary from the number
required for optimum voltage transfer to zero, only increases the
selectivity by a factor of two. This means simply that when the
number of turns in the primary is zero, none of the plate resistance
of the tube is transferred to the secondary circuit, and, of course,
then the selectivity of the secondary is its selectivity when stand-
ing alone. No voltage is transferred under this condition, and so
such selectivity for a single stage is never attained.
There is another variable factor in this transformer, this is the
ratio of L to C. We have already discussed the voltage gain and
selectivity as controlled by the secondary resistance and the turns
ratio and the plate resistance of the previous tube. Calling upon
either experimental proof or mathematical analysis, we can show
that the greater the inductance of the secondary, other factors
being maintained, the greater will be the selectivity of the system,
and that when the secondary is adjusted to give the maximum
amplification, given by
RR, = w2L?
(where w = 2 X resonant frequency) the selectivity will depend
upon the ratio of L to R, (secondary resistance) only and that if
the circuit is adjusted for the greatest amplification for a desired
signal it will also give the smallest amplification to an unwanted
signal (K. W. Jarvis, Proc. I. R. E., May, 1927). Other curves
SELECTIVITY 315
showing the influence of coupling, etc., on selectivity and amplifi-
cation are shown in Figs. 188 and 189.
Relative Amplification
10 5 0 5 10 15 20
K.C. off Resonance
Fic. 188.—Curves showing how selectivity of tuned r.-f. circuit depends
upon frequency.
252. Selectivity—So far we have treated the selectivity prob-
lem in a qualitative way only. We have not said anything about
how much selectivity we needed or desired, or how much we could
get. We stated that increasing the coupling between the primary
and secondary circuits
(decreasing N) of our
tuned transformer in-
creased the resistance
in the secondary
circuit and thereby
decreased the selec-
tivity of that cir-
cuit, and that at the
coupling for maxi-
mum voltage gain
the selectivity was
halved. What does
this mean? How
much selectivity is
needed to prevent
cross-talk between stations 10 ke. apart?
20
15
G
10
: L=205 uh
Rp=8 800
#=8.7
1500 1000 750 600 500
K.C.
Fig. 189.—Relation between mutual inductance
and gain in an r.-f. amplifier.
How much selectivity
316 RADIO-FREQUENCY AMPLIFERS
can be secured? What are the coil characteristics to provide such
selectivity? How much selectivity can be tolerated before side
band cutting becomes audible to the ear? These are questions
that assail every receiving set engineer.
In Chapter VII we discussed the simple series resonant circuit,
and stated that the steepness of its response curve at resonance
depended upon the resistance in that circuit. The greater the
resistance, the duller the curve, and the less difference in voltage,
or current, between the resonant frequency and frequencies off
resonance by say 10 ke. In fact the width of the resonance curve
is a direct measure of the resistance in the circuit. Consider such
resonance curves as that in Fig. 188. If fi and fe are two fre-
quencies on either side of the resonant frequency, f,, such that the
current, or voltage, at these frequencies is 0.707 of the value of
voltage, or current, at resonance, then the following formula is
true:
Here again we run into an expression involving the inductive reac-
tance and the ohmic resistance of the coil. It states that knowing
the value of Lw/r of the coil, we can tell at once how wide the
resonance band is going to be at a point where the current, or
voltage, is 0.707 of its maximum value. (This amounts to a 3 DB
loss in voltge in a single circuit.) How is this useful in our present
problem?
The tuned transformer used in t.r.f. sets, or neutrodynes, or in
any system in which the secondary of a transformer is tuned, may
be considered as a simple series circuit provided we choose the
constants of that series circuit correctly. The resistance of that
circuit will be its series resistance, 7, usually consisting of the
high-frequency resistance of the coil alone, plus the resistance
“reflected ” into that circuit by the transformer. This resistance
considered as shunted across the tuned secondary is the plate
resistance of the tube stepped up by the square of the effective
turns ratio of the transformer, just as in any transformer case.
Now any resistance shunted across the tuned circuit is equiva-
SELECTIVITY 317
lent to a smaller resistance in series with the circuit, and such a
resistance is the controlling factor in determining the width of the
resonance curve. And since selectivity is but a term describing
the width of this resonance curve, we have in our hands all the
necessary facts regarding the voltage amplification and the selec-
tivity of such circuits.
For example, at the coupling between the primary and secon-
dary for maximum voltage amplification the effective turns ratio
is given by
which means that the secondary of the coil will be shunted by a
resistance equal to the plate resistance of the previous tube mul-
tiplied by the turns ratio squared, and this numerically will be
equal to the effective resistance of the tuned secondary, L?w?/r.
Now the secondary effective resistance, being shunted by another
resistance equal to it numerically, becomes half its former value
(two equal resistances in parallel have a resultant resistance of
one-half of one of them). Such a decrease in effective resistance
can also be produced in one other way—by increasing the resist-
ance of the coil to twice its normal value. And what effect has
doubling the coil resistance upon selectivity? Clearly it halves it,
because selectivity may be thought of as proportional to Lw/r and
doubling r halves this factor.
The following example may fix the whole problem in one’s mind.
/ Example 1-13. Consider a coil whose inductance is 200 microhenrys, and
whose resistance r at 1000 kc. is 10 ohms. This coil is to be tuned and fitted
with a primary to work out of a tube whose plate resistance is 12,000 ohms
and whose » is 8. What is the proper turns ratio for maximum voltage
amplification? What is the voltage amplification? What is the width of the
frequency band with and without the tube connected across the primary?
Effective resistance of coil-condenser alone
L2w2 (200 X 10-8)2 (108 X 6.28)?
ro 10
= 158,000 ohms
318 RADIO-FREQUENCY AMPLIFIERS
Width of frequency band when J = .707 maximum value
=f i Sor ellie
sah 92 : | eo By
LOS ero
~ 200 X 10-8 X 6.28 x 108
= 8000 cycles.
Turns ratio for maximum voltage amplification
aise eee SS, COOL:
= = «|: 8-6.
z: 12,000
_ phe 8 X 200 X:10-* x 6.28 X 108
2V r Ry 2V12,000 V'10
= 14.5 times.
Voltage gain
G
Resistance reflected from primary to secondary
= Ly XN?
= 12° 000)>< 135 =9158;000:
New effective resistance of secondary circuit
158,000 X 158,000 L%w?
Bee ee) 79 000 hina eats
158,000 -+ 158,000 eg? ae 5)
New effective series resistance in secondary
L2w? 158,000 X 10
"79 O00) * WE79. 000m
Width of frequency band where J = .707 maximum value
f, Ri 10° X 20
Lw 200 X 10-% x 6.28 « 105
= 16,000 cycles.
=fe—-fi=
This indicates that a voltage gain of 14.5 would be obtained
with such a coil-condenser combination and tube at a frequency
of 1000 ke. and that at a frequency 8000 cycles either above or
below resonance, the current in the secondary circuit, or the voltage
across it, would be 0.707 of its value at exact resonance, This
SUMMARY OF RADIO-FREQUENCY AMPLIFIER PHENOMENA 319
voltage ratio corresponds to a loss of 3 DB, which is not very
great. The selectivity of the transformer and tube, then, is not
very great, certainly not great enough to provide much discrimina-
tion between a station on 1000 ke. and another at 1010 ke. Accord-
ing to the data of Espenshied (Problem 16-12) we need to be
“down” about 40 DB at the frequency of an unwanted station
so that its signals will not interfere.
253. Summary of radio-frequency amplifier phenomena.—A
discussion of voltage amplification and selectivity and of the con-
stant compromise that must be made between them will be found
in Professor Hazeltine’s Patent No. 1,648,808.
The gist of this excellent discussion follows. If the physical
coupling between primary and secondary is close, the turns ratio
between secondary and primary will be assumed equal to VN. We
shall assume that there is no appreciable capacity across the
primary and that the circuit does not tend to regenerate. It has
been properly neutralized.
At resonance at a given frequency the voltage amplification is
given by
Nu
Nee
gr
G
where WN = turns ratio;
# = amplification factor or tube = 8;
r Cr:
= = — = .01667 millimho;
Seer 2a2 °f.
Ga= cS: = 0.1333 millimho,
Ry
7500 ohms
whence fk,
Le?
itp
60,000 ohms.
Using this formula and the special data given on it for a special
set of conditions, the curves in Fig. 190 were plotted, which tell us
320 RADIO-FREQUENCY AMPLIFIERS
a great deal regarding sensitivity, and selectivity as a function
of the turns ratio between primary and secondary. In Fig. 190
we have indicated the several values of N, which is always
equal to the total number of turns divided by the number used as”
a primary.
With N = 1, the voltage amplification at resonance is equal to
16.1 DB (a voltage amplification of about 8). Now the horizontal
axis represents percentages off resonance. That is, if the resonant
frequency is 1000 ke. the point marked ‘‘ 0.10 ” is actually 100 ke.
nD
+
iS
oO
Amplification i
PO
OO OMRRPNON
ry
= 1 Oe OSiee OGIO 4an02 0 .02 .04 .06 .08 .10
Relative Deviation in Frequency
Fic. 190.
(1000 X 0.1) off resonance. If the resonance frequency is 600 ke.
this point will be 60 ke. above or below resonance. At this point,
with unity turns ratio, N = 1, between primary and secondary,
the amplification has only decreased to 6.1 DB, a loss of only
10 DB, 100 ke. off resonance. This is poor selectivity.
When JV is increased to +/2 the amplification has increased
and so has the selectivity. The amplification is now 17.5 DB and
at 100 ke. (for a 1000 ke. resonant frequency) the amplification
is only 3.4 DB.
The amplification soon begins to fall off with increase in turns
ratio (decrease in primary turns) but fortunately the selectivity
SUMMARY OF RADIO-FREQUENCY AMPLIFIER PHENOMENA 321
increases, and so there is a point at which the best compromise
between selectivity and sensitivity can be reached.
The fact that the selectivity changes at each frequency is
-clearly indicated by these figures. At 1500 ke. and a turns ratio
of 4, there is a loss of 25.5 DB at 150 ke. off resonance, but at 550 ke.
the same loss occurs at only 55 ke. off resonance. The response
curve at this frequency would be almost 3 times as sharp as that
at 1500 ke.
The curves show conclusively the fact that there is a certain
turns ratio which produces the greatest amount of amplification,
and that the selectivity increases as the turns ratio is increased.
The best value, according to this discussion, for N is about 2.85
at which the selectivity-amplification ratio is the best. Further
increases in the turns ratio produce too great a drop in amplifica-
tion to be economical. Selectivity must be gained in some other
way. This value of N depends upon coil and tube constants.
Now let us look at these curves with the problem of fidelity in
mind. The response at 5000 cycles off resonance must not be too
much decreased over that at resonance. Thus at 1500 ke. at a
point 0.5 per cent (0.005) off resonance or 7.5 ke. the drop in
amplification with N = 1 is only 0.1 DB. At the most selective
frequency, 550 ke., the loss at 5.5 ke. is only 0.1 DB.
Now the designer of the radio-frequency amplifier has several
factors under his control in addition to the turns ratio. One of
these is the resistance of the coil he uses; another is its inductance.
Decreasing the resistance increases the resonant amplification
but does little to the amplification 10 per cent off resonance.
This amounts to an increase in selectivity. The turns ratio
for maximum amplification at decreased resistance must be in-
creased.
If the impairment of fidelity occasioned by using better coils
must be overcome, a change in turns ratio to give equal fidelity
still gives some increase in sensitivity but with the same selectivity
and fidelity, of course. The turns ratio, N, would be decreased in
this case. The gain in sensitivity, however, would probably not
be worth the effort because of the greater bulk of the coil necessary
to get a much lower resistance. According to Hazeltine, at the
322 RADIO-FREQUENCY AMPLIFIERS
lower broadcast frequencies (600 kc.) fidelity begins to limit the
extent which one can go in the desire for low loss coils. Coils of
greater dimension than 3 inches cannot be used unless fidelity is
permitted to suffer. Since it is rare indeed that a single radio-
frequency amplifier is used, and has become standard practice to
use several stages, each of which in itself has rather poor selec-
tivity, the use of coils of small diameter is almost universal. There
are other advantages as we shall see later.
There is one other variable; the designer can change the ratio
of inductance to capacity. He can use a large inductance and a
small capacity or a small inductance and a large capacity within
the limitation that the same frequency range will be covered.
Doubling the inductance, halving the maximum capacity, and
keeping the high-frequency resistance of the circuit constant—a
difficult if not impossible requirement, and using the proper turns
ratio will increase the amplification at resonance without decreas-
ing the selectivity or fidelity. For example increasing the ordinates
of all the curves in Fig. 190 by 3 DB will give the result of doubling
the inductance, etc.
254. Selectivity to signals far off resonance.—The curves in
Fig. 190 show that to signals far from resonance the amplification
is a negative quantity. This means simply that there is a loss at
these frequencies, and that at the end of several stages signals far
from resonance will be weaker than if only the final stage were used.
This is one argument for careful shielding. The direct pick-up by
the detector circuit of an unshielded set may cause a greater signal
from an unwanted station than would be heard if the preceding
and detector input were carefully shielded. There is then no
direct pick-up, and whatever gets to the detector input must
come through the antenna-ground channel.
The amplification of frequencies 10 per cent off resonance is
practically independent of the ratio of resistance to reactance of
the transformer, and so variations of coil resistance at various
frequencies within the band over which the receiver will be tuned
will not make much difference to signals far from resonance.
255. Use of several stages of radio-frequency amplification.—
The preceding summation of the Hazeltine paper has regard to a
COIL FACTORS 323
single stage only. Such a single stage will not give enough ampli-
fication except over medium distances, and not enough selectivity
to cope with the multiplicity of stations in the United States. It
becomes necessary to use two or more stages. Jn this case, the
voltage gain of the entire amplifier may be obtained by multiply-
ing the ordinates of a single response curve by the number of
identical stages employed. The same result will be secured by
adding the DB gain or loss at each point on the response curve
of a single stage. Increasing the number of stages increases the
amplification at resonance, increases the selectivity and impairs
the fidelity unless the turns ratio between stages is varied so that
constant fidelity results. The data in the table below are from
Hazeltine’s discussion in the Proceedings of the I. R. E. which
gives the significant statement, ‘‘ with constant fidelity the selec-
tivity is increased as the number of stages, whereas with a fixed
number of stages the selectivity can be increased only at the
expense of fidelity.”
With Constant Turns Ratio With Constant Fidelity
Number
of Tur Amplifica-| Total ane Amplifica-| Total
Stages, ee ; tion |Amplifica- Ratio tion |Amplifica-
‘e a!) | per Stage,| tion, N ” |per Stage,| tion,
DB DB DB DB
1 4 24.1 DANY S| eel aerate ee a, aceeeeca ts | een ee
2 4 24.1 48 .2 6.21 23.3 46.5
3 4 24.1 12.2 4.67 24.0 TOL 8)
4 4 24.1 96.2 4.00 24.1 96.3
8 4 24.1 192.6 2.96 236 189.6
16 4 24.1 383.3 2.31 22.8 365.6
Boot rcne te crat ek: 4 24.1 eae | (stare esiewae Nadas sete, <i-s¥aite | levoneae teen
256. Coil factors.—So far as the coil in our radio receivers is
Lw ;
concerned selectivity is proportional to ae the effective turns
324 RADIO-FREQUENCY AMPLIFIERS
Lew :
ratio for maximum voltage gain N = ——=, and the maximum pos-
Vv rRp
1
AVA fe,
The data in the following table give the width of band passed,
maximum voltage amplification, etc., at three frequencies in the
broadcast band using a tube whose uy is 8, and plate resistance is
12,000 ohms, with a coil of 200 uh. In each case the proper turns
ratio is given and the maximum voltage amplification, G, is calcu-
lated for the case in which this turns ratio is used.
Out of such a table may be gleaned several interesting facts.
In the first place the amplification is not constant over the fre-
quency band. In fact in a frequency band that varies from 500
to 1500 ke. (8 to 1 change in frequency) G varies about 2 to 1,
and the turns ratio to produce this gain varies in the same ratio,
and the width of the frequency band at the 0.707 voltage point
differs at each radio frequency.
wp Lw
sible voltaye amplification at any frequency G = 9 x ae x
.
Ren = idth of
J ie eff ; Width o a
Fre r of ie L2w2 Bend iCal N G Width of
quency | Coil r i Nea Band
500 ke. 6 | 104 66,600 4,770 2.4 7.3 | 9,540 cycles
1000 ke. | 10 | 125.6] 158,000 7,950 3.6 | 14.5 | 16,000 cycles
1500 ke. | 15 |125.6] 188,400 12,000 4.0 | 16.0 | 24,000 cycles
257. Turns ratio into detector tube.—A grid leak and condenser
detector has a low input resistance because of the flow of grid cur-
rent. This resistance is of the order of 60,000 ohms into which the
previous r.-f. amplifier must work. The turns ratio, for maximum
voltage amplification, of the transformer coupling the detector and
the previous tube must be somewhat lower than between two r.-f.
amplifier tubes. This type of detector, however, is notoriously
broad in tuning because of this low input resistance, and for this
reason the turns ratio as used in commercial practice is kept fairly
large so that the tuning of the detector does not differ appreciably
from that of the other tuned circuits.
REGENERATION AND OSCILLATION IN R.-F. AMPLIFIERS 325
258. Regeneration and oscillation in r.-f. amplifiers.—All of
the discussion up to this point assumes that the tubes and trans-
formers operated in a stable manner, repeating into the plate circuit
what appears in the grid circuit in amplified form, and not repeating
back to the grid circuit any of this amplified voltage. Practically,
such conditions do not hold. Unless precautions are taken, the
circuit oscillates long before the proper inductance has been added
to the plate circuit to provide a reasonable amount of amplification.
There are two reasons for this unwanted oscillation. One lies
in the uniatentional couplings provided between output and input
circuit, for instance through mutual inductance between the coils,
through capacities which connect the two circuits, and through
other couplings. The other source of coupling is also unintentional
but, unlike those mentioned above, cannot be eliminated. This
second coupling is that existing within the tube, and is due the
capacity between the plate and the grid. So long as the plate and
grid are at different potentials there will be a capacity current,
and since the plate is part of the output circuit, and the grid
is part of the input circuit, what takes place in one circuit has
some effect on what is taking place in the other circuit.
Let us look at Fig. 191 and assume that there is an amplifica-
tion in voltage of 10 in the tube, that is, whatever voltage, H;, is
placed across the input appears as
10 times this value in the output.
The voltage is applied to the input
through the mutual inductance
between the primary, P, and
secondary, S, of the transformer.
Whatever a.-c. current flows in
the plate circuit must also go
through the coil, 7, commonly
ealled a “ tickler.”’ If this coilis Fie. 191.—A simple regenerative
coupled to S in the proper man- circuit.
ner, it will induce a voltage in the
input coil S in such a direction that it will be in phase with the
voltage induced there from P. Suppose the voltage due P is 1.0
volt and that the voltage on S due to T is one-half volt. We now
M Variable
———
326 RADIO-FREQUENCY AMPLIFIERS
have not 1 volt on the grid of the tube but 1.5 volts. This is
what is known as regeneration; part of the output voltage is fed
back to the input and in phase. When the feed-back voltage is of
the correct magnitude and phase, we may remove the input voltage
due the primary, P, and a.-c. currents will still flow because what-
ever came from P originally has been amplified in the tube and
fed back to the grid where it is amplified again, and again returns
to the input. In other words the tube oscillates; it supplies enough
energy from the B battery to wipe out all the losses in power in
the circuit.
Let us look at the phenomenon of regeneration in another way.
Suppose the input circuit to the tube is tuned. The current in the
tuned circuit is controlled by the resistance in that circuit—pro-
vided a constant voltage is impressed from P. If, now, we decrease
the resistance in the circuit the current increases, and the voltage
across the circuit (and hence on the grid) increases.
Now suppose the voltage across the input is increased by means
of the tickler coil, 7. This produces exactly the same result as if
the resistance in the tuned circuit were decreased. We may
express what has happened by stating that feed-back, if in proper
phase and magnitude, may introduce a negative resistance into the
tuned circuit. This negative resistance added to the already
existent positive resistance decreases the total resistance’ there.
When the tube feeds back sufficient negative resistance so that the
entire resistance losses in this circuit are wiped out, the system
maintains itself in a state of continuous oscillation requiring no
additional a.-c. energy from without, and capable of supplying
considerable a.-c. power to some external circuit coupled to it.
Such a feed-back of voltage from the plate to the grid circuit
may take place through desired coupling—as in case of the tickler
feed-back—or through unwanted coupling, as mentioned above.
The grid-plate capacity is the most prolific source of trouble from
regeneration, because the voltage fed back from this small inter-
element capacity may be of the proper phase and magnitude to
cause not only regeneration but oscillation as well.
Section 245 mentioned the fact that the input impedance of a
vacuum tube is not a pure capacity, but that it may be a capacity
REGENERATION AND OSCILLATION IN R.-F. AMPLIFIERS 327
plus a resistance which may be either positive or negative depend-
ing upon the load in the plate circuit. If the load is a positive reac-
tance, an inductance, the input resistance of the tube may be
negative, and so the voltage fed back there from the plate circuit
through the grid-plate capacity will be in phase with the voltage
already appearing there. Regeneration takes place. If the resist-
ance of the input circuit is sufficiently negative the circuit may
oscillate.
If the load in the plate circuit is a negative reactance, a capac-
ity, the input resistance of the tube will be positive, and will shunt
the input circuit. If this circuit is a coil-condenser combination
tuned to resonance with some incoming voltage, the positive resist-
ance of the grid-filament circuit of the tube will be placed across
this tuned circuit and of course will decrease its selectivity.
The input impedance of a tube, then, depends upon the plate
load. Any change occurring in the plate circuit is repeated back
to the input grid circuit through unwanted couplings, usually
through the grid-plate capacity, and may cause either regeneration
due to an inductance load or degeneration due to a capacity load,
which produces a positive input resistance.
If the load in the plate circuit is resistive or capacitive the tube
and circuit cannot regenerate or oscillate. Signals may even be
weaker. In Fig. 192 if the plate
circuit is exactly tuned to resonance
with the frequency of the signals com-
ing from the input circuit, the plate
load is resistive—the impedance of
an anti-resonant circuit at resonance
is resistance only—and so the circuit =c +B
will not oscillate, a fact that is often Fy¢. 192 —Tuned plate-tuned
overlooked. If the condenser ca- grid circuit.
pacity is greater than that required
for resonance, the load in the plate circuit is essentially ca-
pacity and the circuit cannot oscillate. If, however, the con-
denser is reduced, so that the circuit begins to act like an
inductance, the circuit begins to regenerate and finally oscillations
may begin. If the plate circuit capacity remains fixed and the
328 RADIO-FREQUENCY AMPLIFIERS
frequency of the input signals is varied, a frequency will be reached
which makes the plate circuit resonant. For frequencies above
this critical value the plate circuit is capacitive, and no regen-
eration takes place. For frequencies lower than this critical fre-
quency the load is positive (inductive) and regeneration begins.
259. Losses.—We are faced with the predicament of having
to use inductance in the plate circuit in order to transfer energy
from one tube to another and of having to cope with unwanted
regeneration or oscillation due to this positive reactive plate load.
What can be done about it?
There are several ways in which receiver designers have tried
to solve the problem. The negative resistance introduced into the
Rs
Fic. 193.—Series (R;) and parallel (R,) Fre. 194.—Potentiometer method
losses added to prevent oscillation. of controlling oscillations.
grid circuit by the plate-grid capacity may be reduced by adding
resistance to this circuit. This resistance may be added in series
with the tuned circuit, as R, Fig. 193, or across it, as R,. Oscilla-
tion can take place only when the total losses in this tuned circuit
are wiped out by the feed-back voltage, and so, having the losses
under control of the operator, the circuit can be maintained in a
stable condition at all times.
Another method is the potentiometer method, as in Fig. 194,
where a potentiometer is across the A battery. The grid return
is connected to the movable arm of this unit. When oscillations
start, the arm is thrown to the positive side of the potentiometer so
that considerable grid current flows. This lowers the in put resist-
ance of the tube and decreases the resistance across the tuned
BRIDGE SYSTEMS 329
circuit, which is the same as adding series resistance to it. And
so oscillations cease again.
These methods decrease the amplification at the same time they
decrease the oscillations. The potentiometer method also con-
sumes considerable B battery current because of the high plate
current when the grid is positive, and is none too conservative of
tube life. At the same time the selectivity of the circuits decreases
at a rapid rate.
Other methods, which merely decrease the amplification to the
point where the feed-back voltage is not sufficient to produce oscil-
lations, include reducing the A battery filament current, or the B
plate voltage. Both of these methods increase the plate resistance
of the tube and therefore a smaller a.-c. voltage appears across the
plate load.
Another method that has come into common use is the so-
called ‘“ grid suppressor’? method. It involves placing resistance
between the tuned circuit and the grid of the tubes. The values of
the resistance may be from 300 to 1000 ohms.
260. Bridge systems.—There is another large class of circuits
in which unwanted feed-back is fought in another way—a way
that seems more scientific to many engineers, although it may be
not a great deal more effective. These circuits are the “ bridge
circuits’ of Hazeltine, Rice, Hull, Ballantine, Hartley, and
several others.
The Rice circuit is the simplest to understand. It appears in
Fig. 195. It involves tapping the input coil in the exact center,
and connecting a ‘“‘neu-
tralizing’’ condenser
from the plate of the
tube to the bottom
of this coil. If the
input coil is tapped
at the exact center,
and if the neutraliz-
ing condenser has the Fig. 195.—Rice neutralized circuit.
same capacity as the
grid-plate capacity, for every voltage fed back through the
330 RADIO-FREQUENCY AMPLIFIERS
latter capacity, of the proper phase to cause regeneration—due
to the inductive load in the plate circuit—there will be an equal
and opposite voltage fed
back through the neu-
tralizing condenser. The
equivalent bridge circuit
is shown in Fig. 196. If
the coupling between the
two halves of the input
circuit is perfect and if
the grid-filament capac-
ity and the filament-
plate capacity (in dotted
lines) are equal, the
Fra. 196.—Equivalent bridge circuit of Fig. 195. bridge will be balanced
at all frequencies. Ac-
tually these conditions do not exist and so there may be some
regeneration, or even degeneration in a given circuit.
261. The neutrodyne.—The neutrodyne of Hazeltine gets the
neutralizing voltage from the plate voltage instead of the grid
voltage as shown in Fig. 197. The
Roberts system, Fig. 198, also uses Cin
plate circuit neutralization. Other
systems are in use, but in general
they are more complex than these P
illustrated here.
The Rice circuit has the advan-
tage that the circuit is complete in
itself and no wires need to go to any
other circuit for neutralizing volt-
ages. The plate circuit load may be
placed at some distance from the
amplifier tube itself. It has the
disadvantage that half the input
voltage is not usefully used, that is, it is not applied to the
grid-filament path of the tube. It also has the disadvantage
that both sides of the tuning condenser are above ground potential,
Fig. 197.—Hazeltine neutralized
circuit.
NEUTRALIZING BRIDGE CIRCUITS 331
one side being connected to the grid and the other to the neutraliz-
ing condenser. Some trouble with “ hand capacity ” will be expe-
rienced in using this circuit unless precautions are taken to use a
non-metallic shaft on the condenser.
The Rice circuit is troubled with parasitic oscillations, that is,
oscillations at some other frequency than those determined by the
capacity and the inductance of the tuned circuit. For example, in
a Rice neutralized amplifier operating on broadcast frequencies
oscillations frequently take place on a wavelength of about 75
meters, corresponding to the induc-
tance of half the input coil and
the capacity across it due the grid- N
filament capacity of the tube, wiring, of
etc. The other half of the input coil i
may be thought of as a tickler.
A high loss put into this oscil- ie
lating circuit, as R in Fig. 195, will tig. 198—*«Roberts” neutral-
stop all such oscillations. Such a loss ized circuit.
may be a high resistance, 500 ohms
will do, a choke coil, or an anti-resonant circuit tuned to the
offending frequency.
262. Neutralizing bridge circuits.—A single neutralized ampli-
fier is often placed ahead of a regenerative detector by experi-
menters. In such cases it is only necessary to make the detector
oscillate and to so adjust the neutralizing condenser that tuning
the input to the r.-f. amplifier does not throw the detector out of
oscillation. When such a condition exists, the r.-f. amplifier is
independent of its following circuit, and so the detector and the
amplifier may be tuned separately to the same frequency without
disturbing noises. Actually it is practically impossible to neutralize
an amplifier to the point where it will not throw the detector out
of oscillation at some frequency. This is because of other couplings
that exist in addition to the plate-grid capacity.
If such couplings exist their presence will be indicated by the
following test. Make the detector oscillate and pick up a broad-
casting station carrier. This will cause a distinct beat note to
appear in the head phones or the loud speaker. Now vary the r.-f.
332 RADIO-FREQUENCY AMPLIFIERS
amplifier tuning condenser through resonance with the detector.
The beat note will now change. If there are no couplings aside
from the grid-plate capacity the tone of the beat note will change
to a maximum, or minimum, pitch and then return to its original
value. When the tube is properly neutralized this beat note will
not change in tone. But if unwanted couplings exist due to wiring,
or capacities between plate and grid apparatus, the beat note will
have a double hump in it.
The experimenter should not take the exact neutralizing of his
amplifier too seriously. All that is really desired is a value of
capacity such that oscillations will not take place in the rf.
amplifier at any frequency within the band that will be tuned over.
As a matter of fact a slight misadjustment may increase the strength
of the signals.
If an oscillating detector is not present, another method may
be used which leads to the same result. The filament of the tube
to be neutralized is not lighted. A buzzer-modulated signal is
picked up and the neutralizing condenser in this tube’s circuit is
adjusted until the note of the buzzer as heard in the head phones
or loud speaker is a minimum. Then the tube is lighted and the
next stage is neutralized in the same manner.
Such a method is faulty in that the plate-grid capacity of a tube
is not the same lighted as unlighted, but practically the method
leads to stable amplifiers, and this is the ultimate object of
neutralization.
263. Filtering r.-f. circuits—Oscillation is the most serious
difficulty which amplifier designers run into. It is caused, as
stated above, by coupling part of the plate a.-c. voltage back to
the input of the tube. When occurring through wiring, or faulty
layout of apparatus, or from one coil to another, it is unpardonable,
because it shows evidence of poor design. Let us consider the
circuit in Fig. 199. The r.-f. currents should follow the dotted
lines, and should go nowhere else. If they do they are sure to get
mixed up with similar r.-f. currents from other stages of the
amplifier, and thereby cause unwanted coupling. This is analogous
to the audio-frequency amplifier problem of keeping a.-c. currents
where they belong and out of external circuits.
FILTERING R.-F. CIRCUITS 333
Filtering of all B and C leads will keep the a.-c. currents in
their proper places in the circuit and will keep them from becoming
sources of unwanted coupling with other parts of the amplifier.
Such a filter may consist of a 5000-ohm resistor in series with the
plate battery leads and a fairly large condenser in shunt as shown
in Fig. 199. A capacity of 100 mmfd. at 1500 ke. has a reactance
of 10° ohms and is much to be preferred to a large paper condenser
of perhaps 1.0 mfd. capacity which may have considerable induc-
tance in it. Some condensers made of large sheets of paper rolled
up together have such
an inductance that they
present a very high anti-
resonant reactance at the
higher radio frequencies.
For this reason a small
mica condenser of 0.01
or 0.001 mfd. capacity
will provide good by-
passing if the series im-
pedance is fairly high.
Ballantine cites the
case of two No. 18 wires igh Impedance Series Path
two inches apart that Fig. 199.—Proper use of filtering to keep r.-f.
have a reactance of 5.8 currents where they belong.
ohms per foot. A ground
wire carrying r.-f. currents and near a grid wire also carrying r.-f.
currents or of the proper phase may provide sufficient coupling
between circuits to cause trouble from regeneration. No a.-c.
currents should be permitted to flow through the filament circuits,
or the metallic shields if such are used. Shields should be
grounded at only one place, to avoid circulating currents in them.
Magnetic coupling from a plate to a grid coil is a prolific source
of unwanted coupling. One method of avoiding this difficulty is
to use coils in large and heavy metallic containers which are
grounded. Any magnetic field from the coils which would ordi-
narily become mixed up with similar fields from other circuits (and
thereby induce unwanted voltages in them) induce voltages in the
Low Shuntpath
d.c.
334 RADIO-FREQUENCY AMPLIFIERS
grounded shield instead. It must be remembered that the induced
voltage is capable of setting up an r.-f. current in this shielding
and that the J?R loss in power in the shields must be supplied from
the power in the coils themselves. This results in an equivalent
increase in the resistance of the coils inside the shield and a decrease
2.2
in their inductance. This, of course, decreases the coil’s
factor with consequent decrease in both amplification and selec-
tivity. When coils are to be shielded they usually have small diam-
eters and small fields. This construction minimizes the increase in
resistance and decrease in inductance.
264. Use of screen-grid tubes as r.-f. amplifiers.—The screen-
grid tube will probably remove the necessity for neutralizing radio-
frequency amplifiers by removing the source of trouble—the grid-
plate capacity. At any rate it is possible to build tubes of this
type with such low values of capacity that considerably more
voltage amplication can be secured from them without neutralizing
than can be obtained from three-element tubes by means of the
accessory apparatus already described. It must not be supposed
that this new tube—which came into common use in 1929—can
be used without danger from regeneration or even oscillation.
Any tube with considerable amplification will oscillate provided the
circuit constants are correct. If the feed-back due to plate-grid
capacity has been eliminated, it does not follow that one can use a
circuit set-up in which there may be considerable feed-back from
other sources.
Methods of using this new tube will be discussed in Chapter
XV. The voltage amplification possible with a single stage may be
seen in Fig. 182.
CHAPTER XIV
DETECTION
SUPPOSE we have received a signal and have amplified it in a
radio-frequency amplifier. How may it be detected, or demodu-
lated so that it can be put into an audio amplifier and then a loud
speaker?
Up to the present time we have considered the applications of
the vacuum tube which call for its operation on a straight part of
its characteristic where little distortion takes place. We have
considered the tube only as an amplifier. We shall discuss now the
uses for the curved part of the characteristic.
Tubes act as amplifiers either with or without distortion and
as detectors and modulators in which distortion is the essential
feature. The latter uses of the tube require a curved character-
istic. The output no longer is an exact replica of the input.
265. Distorting tubes.—In Section 188 we were able to calcu-
late the amount of distortion (second harmonics) that resulted
when the tube was worked on a curved part of its characteristic,
and found that in the distortion process a certain amount of d.-c.
current was generated. In other words a pure sine wave voltage
put on the grid-filament input of a tube resulted in an output
current or voltage composed of not only the frequency that was put
on the input but some additional frequencies and some additional
d.-c. current as well.
If we desire to get d.-c. current from an a.-c. voltage, the tube
acts asa rectifier. If we desire to get audio-frequency voltages from
a modulated r.-f. voltage, the tube acts as a detector. If we desire
to mix two frequencies, say a low audible frequency with a high
or radio frequency, we put them both into a modulator. All of
these uses require a non-linear characteristic. Detectors and
335
336 DETECTION
modulators have three elements just as amplifier tubes have.
Tubes designed for rectifiers have only two elements, plate and
filament.
266. Modulation—Consider the circuit in Fig. 200. If the
high-frequency generator is turned on and the low-frequency
generator is shorted, high-
frequency currents will
flow in the antenna. Their
amplitude will depend
upon the amplifying ability
of the tube and the am-
plitude of the applied grid
voltage at this frequency.
If the grid voltage is of
constant amplitude and
Fig. 200.—A simple modulator. frequency, the plate cur-
rent variations and hence
the antenna current variations will be of constant frequency and
amplitude. Now let us turn on the low-frequency generator. The
amplitude of the antenna current will vary and if the proper
relations between the plate current and the grid voltage are
satisfied the amplitude of the high-frequency antenna current will
vary at the frequency of the low-frequency generator.
An idea of how the antenna currents look before and after
modulation may be seen in Fig. 201. The high frequency is called
the modulated or carrier frequency, and the low frequency is called
the modulating or side-band frequency.
Let us call the maximum amplitude of the antenna current B,
and its frequency f;. Then the current at any instant will be
t= Bsin2 af in which the “27f” takes the place of the
phase angle 6 but expresses the same thing exactly, namely the
amount of time that has elapsed since the beginning of the cycle.
Now instead of a constant maximum amplitude B let us vary this
amplitude above and below B at some rate, say in the form of a
sine wave. The maximum amplitude is no longer constant, but is
equal to B + A sin 27fmé in which fm stands for modulating fre-
quency just as fc indicates carrier frequency. The current at any
MODULATION 337
instant is a function of several variable factors, and may be
expressed as 7 = (B + A sin 2afnd) sin 2 aft.
This process whereby a high frequency is varied in amplitude
by a lower frequency is called modulation. The system outlined
above is called grid-circuit modulation. Plate-circuit modulation
is explained in Section 363.
The tube in such a process is called a modulator. Once the
high-frequency wave is modulated, it acts as a carrier for the low
frequency and wherever it goes it takes the modulating frequency
with it.
During Modulation
Fig. 201.—Unmodulated and modulated wave.
The depth to which the high frequency is modulated depends
upon the relative maximum amplitudes of the two frequencies.
If they are equal, the wave is said to be completely modulated, and
the “‘ percentage modulation ” is 100.
267. Percentage modulation.—If the two peak voltages are not
the same, the high-frequency wave will not be completely modu-
lated, and the modulation will be less than 100 per cent. In
broadcast transmission the modulation rarely exceeds 90 per cent.
The greater the modulation percentage the further will the signals
from a given station be heard and the greater will be the distortion
arising from demodulators or detectors following a “ square law.”
In 1929 the use of “linear” detection became prevalent and it
338 DETECTION
became possible to receive 100 per cent modulation without the
“square law ”’ detection distortion.
In Fig. 201 is the carrier before and after modulation. The
percentage modulation is defined as the ratio between A, the peak
current of the modulating frequency, and B, the peak current of the
non-modulated carrier. A glance at Fig. 201 will give a good idea
of what is meant by the expression. When 100 per cent modula-
tion is effected, the values of A and B are equal.
A
Percentage modulation = M = B X 100 per cent.
268. Demodulation.—If such a modulated wave is turned into
a ‘ demodulator,” the side-band frequencies can be got back. A
demodulator, or detector, acts as though it were made up of two
filters, of which one will not pass the high or carrier frequency
and the other will not pass the low or modulating frequency. In
the demodulator the two frequencies are separated.
A modulator, then, is a device for combining two frequencies.
A demodulator, or detector, is a device by means of which we get
back from the radio-frequency carrier the modulation or intelli-
gence carrying frequencies.
269. A simple detector.—Let us consider the case of a device
which has a response characteristic like XYZ in Fig. 202. When
the voltage across it is as large as 4 volts, current begins to flow
through it, and from then on the current is proportional to the
voltage. Below this value, no current will flow. Now suppose we
put an a.-c. voltage on this device. If the average value of the
a.-c. voltage wave is at the point A, the same current flows when
the voltage is positive or negative—the current wave looks exactly
like the voltage wave, just as in the distortionless amplifier. But
suppose the average value of the sine wave is at the point B. Then
current flows only on the positive halves of the cycle. Such a
device is a rectifier or detector and if the operating point is at the
exact place where the current begins to flow (as at B), perfect
rectification will result.
Now let us consider an unmodulated wave. When it passes
into a perfect rectifier, current flows during the positive half-cycle
A SIMPLE DETECTOR 339
only. Then, if the normal steady current is zero milliamperes,
when no a.-c. voltage is applied, this current will become some
positive measurable value when the a.c. is applied because, though
a meter cannot follow these spurts of current, it will take up some
average value between the peak of the spurt and the zero value—
which is the same as the normal no-signal value of the current.
Average Value=0
A
Average Value=1.5
B
Se
A
Fic. 202.—Rectification takes place when the operation is at (Y) but not at
(A). Thus the plate current in (A) is similar to the input voltage while at
(B) the output differs from the input.
In such a detector there is an increase in current when a.n€.
voltages are applied. Now if the a.-c. voltage is modulated as in
Fig. 203, the average value of detector current goes up and down
in accordance with these modulations. This varying average value
is the useful part of detection since it has the same form as the
original modulating voltage and in a distortionless system is exactly
340 ‘DETECTION
proportional to this voltage. This varying value occurs at an
audible frequency.
270. The plate circuit detector—The simplest tube detector
is the plate circuit or C bias detector, that is, a rectifier which
operates upon a curved part of the grid voltage plate current
characteristic curve of a tube. It is not a perfect rectifier, but as
greater and greater voltages are placed upon it the positive halves
of the current waves are much greater than the negative waves,
and so the average
value of plate current
due to rectification
increases a corre-
sponding amount.
In this process,
there are three points
tonote. First the very
rapid radio-frequency
variations in input
voltage modulated at
an audio rate. Each
audio cycle is madeup
of thousands of radio-
Fia. 203.—Rectification taking place about a non- frequency cycles—
linear part of a characteristic. only a few of them are
shown in the diagrams,
for simplicity. When this voltage is placed upon a curved part of
the plate curve, the difference between the positive and negative
current waves takes the form of the audio variations. Because
these positive and negative portions differ in amplitude, the average
between them is not zero and therefore the plate circuit con-
tains a current of the frequency of the audio signals. In this process
there is an increase in average d.-c. plate current whether the input
wave is modulated or not. The plate current meter cannot follow
the change in audio plate current and does not indicate whether or
not the carrier is being modulated.
271. Detection of modulated wave.—Suppose for example we
connect antenna and ground to the grid and filament input of
Output
DETECTION OF MODULATED WAVE 341
a tube. A nearby station is putting into the ether an unmodu-
lated wave. As soon as we tune to this station’s frequency a
voltage is developed across the input circuit to the tube; this a.-e.
voltage fluctuates the grid voltage, and a change takes place in
the average value of the plate current. If the station is powerful
enough, and if it is modulated with a key—that is, if its antenna
current is started and stopped in accordance to some code—we can
use a sensitive relay in the plate circuit of our detector and either
read the signals directly from the relay or operate a telegraph
sounder with it, or light a lamp, or fire a gun or do anything else
which it is required to control by radio. If the transmitted fre-
quency is 1000 ke. it may be necessary to tune the input of the tube
to this frequency. A pair of telephones in the plate circuit of the
detector, in place of the relay, will indicate by means of clicks when
the transmitter started and stopped the antenna current but would
not give off any sound in the middle of dots and dashes. If, how-
ever, we operated a buzzer with the relay we could read the signals
by the audible sound of the buzzer.
We can get around the difficulty of needing a receiving buzzer
by having a buzzer or “‘ chopper” at the transmitting station.
Now when the key is pressed, the modulated 1000-kc. wave is sent
into the ether. If the modulator tone, say 1000 cycles, has a
maximum amplitude equal to the maximum value of the 1000-ke.
voltage, the antenna current will be doubled 1000 times a second
when these two voltages will be in phase. At 1000 other instants
the two voltages will be out of phase and the antenna current will
be reduced to zero. Across the receiving tube is a 1000-ke. voltage
broken up into 1000-cycle sections—modulated as we say.
Now if we place a pair of telephones in the plate circuit of the
detector tube, it will offer a certain amount of impedance to these
1000-cycle sections of 1000-kc. currents. A voltage will be built
up across the telephones, and our ears will tell us that 1000-cycle
signals are being received. The d.-c. plate current meter will still
indicate an increase in average plate current when the transmitter
key is pressed, but of course its needle cannot follow the 1000-cycle
variations. Neither will a relay indicate that the 1000-ke. input
is modulated, because it is too sluggish to follow the variations,
342 DETECTION
but it will indicate the average of each section and will close when
the key is pressed. The telephones, however, will respond to
frequencies as high as 10,000 cycles per second.
If, now, we use a microphone at the transmitter instead of a
constant peak amplitude source of tone, like a buzzer, and talk into
it the voltage variations impressed on the 1000-ke. voltage may be
very complex, perhaps something like Fig. 203 for example.
Telephone receivers in the detector circuit will respond to the
sections of 1000 ke. but since they are not symmetrical, like a pure
sine wave, the receivers will not give off a pure tone. They will
give off a note or sound corresponding to what was put into the
microphone. If, in the process of detection, other audio frequencies
are generated distortion results, because the audio components
heard in the telephones are no longer exact replicas of what was
put into the microphone. Fortunately these additional frequencies,
the strongest of which are of double the original audio tone, are of
small amplitude provided the detector is properly operated, and
so the tone as heard from the telephone sounds exactly like that
put into the microphone.
The process of detection by means of the bend in the plate
current curve is essentially one of distortion, in which the r.-f.
wave is distorted and out
of which we get the audio
wave by placing some
audio-frequency imped-
nea C poe ance in the plate circuit
of the detector. Usually
the plate circuit has a
a cee ae low impedance to the r.-f.
I I! a.-c. plate currents, ‘that
Fig. 204-—To keep radio-frequency currents ae they me by-p assed as
from the output, a low impedance path (C) im Fig. 204, and some-
and a high series impedance (L) are used. times an additional pre-
caution is taken to pre-
vent any r.-f. voltages from being built up across the audio
impedance in the plate circuit. For example, in Fig. 204 a
choke, L, is used.
L
THE VACUUM TUBE VOLTMETER 343
272. Conditions for best detection.—There are several variable
factors in such a detector circuit as shown in F ig. 204. One is the
grid bias, E., and the other is the plate voltage. Some combina-
tion of these two voltages will create the greatest amount of a.-f.
voltage across a given impedance when a given r.-f. voltage is put
on the grid. The C bias is governed to some extent by the ampli-
tude of the r.-f. voltages to be put on the grid of the tube.
If an r.-f. voltage of 1 volt maximum is to be impressed on
the grid of the tube, a C bias of slightly over this value will take
care of the signals and offer a certain margin of safety. The prob-
lem then is to find the plate voltage that will put the operating
point at a place of great sensitivity. Values for an ordinary tube of
the 201—A type are about minus 4.5 with a plate voltage of 45
and minus 9 with a plate voltage of 90.
In such a detector the greater the r.-f. voltage the greater the
change in average plate current and the greater the audio signal
in the output load. With a given r.-f. voltage the audio signal will
be proportional to the amount of modulation at the transmitter.
That is, if the power from the radio oscillator at the transmitter is
constant but the voltage coming from the microphone varies, the
audio notes at the receiver should vary in exact proportion to the
microphone notes.
Otherwise there is distortion. The audio notes with a fixed
modulation vary approximately as the square of the r.-f. voltage
when the latter are small. That is, doubling the radio-frequency
voltage across the antenna will result in multiplying by 4 the
audio tones from the detector. Actually the square law holds only
over the certain small part of the plate current curve, for example:
for low input signal voltages. The law is less than a square func-
tion for large input voltages. The change in d.-c. current (increase)
is independent of the modulation and a meter in the plate circuit
will not tell whether or not the r.-f. input is being modulated.
273. The vacuum tube voltmeter.—The detector may be cali-
brated and used as an a.-c. voltmeter. Such vacuum tube volt-
meters are useful at all audio and nearly all radio frequencies, and
can be made to read d.-c. voltages. The range of voltages that can
be measured is very large, the upper limit being the voltage the
344 DETECTION
tube can stand without breaking down, and the lower limit depend-
ing upon the sensitivity of the indicating instrument.
The principle of such devices is simple. The operating point is
chosen, by adjusting the C bias and the plate voltage, so that it is
on a point of considerable curvature. When an a.-c. voltage is
put on the grid, rectification takes place in the plate circuit, and
the d.-c. part of the rectification product is read on a d.-c. instru-
ment. If the voltmeter is properly biased, its input resistance is
very high and it takes so little power from the device whose voltage
is being measured that it may be considered as having no effect
upon the circuit. ‘
The choice of C bias depends upon the input voltages to be
measured. Let us suppose we are to measure a peak voltage of
5 volts. Clearly the C bias cannot be less than this because of the
decreased input resistance when the grid draws current and the
effect of such a meter upon the circuit under measurement. The
C bias would be some value over 5 volts, 6 for example. The next
step is to fix the plate voltage. This is determined by the range in
input volts to be measured, and the kind of instrument used to
read the d.-c. current. For greatest accuracy EH, should be such
that the greatest deflection of the current meter will be obtained
by the given input voltage. In general a voltage range of about
5 to 1 is all that can be read with an ordinary voltmeter, that is,
from about 0,5 volt to 3.0 volts. The problem then is to choose
a plate voltage that will enable the desired range to cover com-
pletely the scale of the meter being used.
For measurements of average voltages, say up to 10 or 15 volts
and as low as 0.5 or 1.0 volt, a microammeter reading 200 micro-
amperes and costing about $35 is a good instrument. A small
laboratory model of milliammeter reading 1.0 or 1.5 milliamperes
can be used although the accuracy of measurement will not be so
great as with a more sensitive instrument.
274. Adjusting a voltmeter.—What is desired is the greatest
change in plate current with a given a.-c. voltage. In Fig. 205
are some curves taken with a 3-volt, 60-milliampere tube, showing
the plate current at various values of a.-c. grid voltage and at
various values of plate voltages. The change in plate current, that
ADJUSTING A VOLTMETER 345
is, the value with a.-c. input voltage minus the value without input,
is plotted and shows the futility of using plate voltages greater than
35 volts when sensitivity is the criterion.
22. 24 26 28
E Volts
Fig. 205.—Calibration and circuit of vacuum tube voltmeter
Because the tube takes some plate current even when there is
no a.-c. grid voltage, part of the scale of the d.-c. meter is taken
up with this steady read-
ing. This reading canbe o—™
balanced out by means
of the zero adjuster on
the meter, or by using —
another current through
the meter in such a di-
rection that the original
plate current is “‘ bucked E
out.”” The whole meter nt
scale is available then y,¢, 206—Method of balacing out the steady
for plate current changes plate current from the indicating meter.
occurring under input
grid voltage excitation. Such a balancing out voltage may come
346 DETECTION
from an additional battery and adjustable resistance, as in Fig. 206,
or from the voltage drop across the tube filament as in Fig. 205.
Experiment 1-14. Use of the Vacuum Tube Voltmeter.—There are two
types of vacuum tube voltmeter: the C bias type and the grid leak and con-
denser type. The latter is more sensitive but draws current from the device
whose voltage is being measured. Both types should be experimented with.
They are the most versatile and useful of all radio instruments.
Connect up a C bias voltmeter using a 0-5 milliammeter in the plate cir-
cuit, about 45 volts on the plate, and add sufficienct C bias to decrease the
plate current to nearly zero. Then use a more sensitive plate current meter,
and a bucking battery to reduce the deflection to zero. Calibrate the meter
by putting known currents through known resistances at 60 cycles or at any
radio frequency. With 45 volts on the plate and about 9-13 volts C bias with
average tubes, voltages as low as 1.0 will give a good deflection and peak
voltages up to about 7.0 can be read on a meter reading about 1.0 milliampere.
Fairly accurate calibration may be performed by using the voltages avail-
able from a filament current transformer, that is, 1.5, 2.5, and 5.0, and of course
combinations of these voltages depending upon how the windings are con-
nected together.
Connect the voltmeter across the tuned circuit of a radio-frequency stage,
or across the coil-condenser combination that can be coupled to an oscillator.
Tune the condenser and note how the voltage across the coil increases through
resonance. Insert some resistance into the tuned circuit and repeat. Note how
much broader the resonance curve is and how the minimum voltage has
decreased.
Connect the voltmeter across the resistor of a resistance-coupled amplifier
tube—inserting a large capacity between the plate terminal of the amplifier
tube and the grid of the voltmeter so that the d.-c. voltage across the resistor
will not bias the grid of the voltmeter. Apply a known voltage to the input
of the amplifier and measure the output voltage. Then change the plate
resistor and again measure the output voltage. Plot a curve showing amplifica-
tion against plate load resistance.
Place the voltmeter across the secondary of an audio transformer and apply
a known voltage in series with a resistance of about 15,000 ohms and the
primary. Measure the turns ratio of the transformer at 60 cycles or some other
frequency by measuring the secondary voltage.
These are but a few of the many experiments that can be performed with
the vacuum tube voltmeter. It can be used to measure field strength of dis-
tant transmitters, resistance of coils, amplification of amplifiers, resonance
curves, frequency characteristics of amplifiers, etc.
275. D.-c. plate current as a function of a.-c. grid voltage.—
The vacuum tube voltmeter is really a C bias or plate circuit
DETECTION IN A RADIO-FREQUENCY AMPLIFIER 347
detector, and the change in plate current is a function of the a.-c.
voltage on the grid. It is only necessary to calibrate the detector,
using any source of a.c. and any standard a.-c. voltmeter as a
standardizing voltage, or a known current can be passed through
a known resistance and the voltage drop used for calibration.
The detector in one’s radio is also a vacuum tube voltmeter
although it is not so calibrated. If a sensitive meter, say reading
up to 5 milliamperes, is placed in the detector plate circuit of any
radio receiver, changes in the reading will be noted when a strong
signal is tuned in. If the detector is a grid leak and condenser type,
the reading will decrease. If greatest sensitivity is desired, the
steady no-signal current may be balanced out and then a sensitive
microammeter may be used. The changes in this meter reading
may serve as a measurement of fading, signal strength, etc. No
change will occur unless the r.-f. voltage on the input to the tube
changes. Modulation will not cause any change unless the trans-
mitter is over-modulated. Generally speaking an r.-f. signal at
ordinary modulation percentages causing a change in detector plate
current of 100 microamperes will, with two stages of audio ampli-
fication, deliver a good loud speaker signal.
276. Detection in a radio-frequency amplifier.—Because the
r.-f. stages of a receiver are biased, it often happens that some
detection takes place in one or more of them, probably in the first.
What happens is something like the following: the first tube is so |
biased, and may have such a low load impedance in its plate cir-
cuit at the frequency under consideration that the operating point
is on a part of considerable curvature. Now a strong signal comes
in, a large a.-c. voltage is impressed on the r.-f. grid and detection
is the inevitable result. A pair of receiving telephones in the
plate circuit of this tube would have an audio-frequency voltage
across them, due to the rectified voltage, and would give an
audible response.
For example suppose the receiver is tuned to a frequency of
600 ke. This means that the load impedance in the plate circuit
of the r.-f. amplifiers will be high at 600 kc. but low to all other
frequencies. The tube will have a curved characteristic to any
signal of higher frequency than this. A powerful local station on
348 DETECTION
some other frequency puts a strong signal on the grid of the r.-f.
tube, and the rectified voltages modulate the r.-f. voltage of the
600-kc. wave so that what gets into the second r.-f. stage is a
600-ke. signal modulated with what is going on at the studio of
the other station. The modulation of the distant station may be
inaudible.
A wave trap tuned to the offending local station is a good
remedy for such trouble.
277. Grid leak and condenser detector.—In the plate circuit
detector, we may look upon the signal as having first been amplified
by the tube and then as going through the detection process when
it reaches the plate circuit. There is little amplification in such a
tube at r.f. because of the low plate circuit resistance at r.f. There
R is another type of de-
tector more complex in
theory and more sensi-
C tive which is in more
AF. Output Common use. This is the
familiar grid leak and
R.F. Input an | condenser detector shown
oo in Fig. 207. In this case
Fie. 207.—Grid leak-grid condenser detector. We nay think of the rf.
signal as going through
the demodulation or detection process in the grid circuit of the
tube and then having the resulting audio tones amplified in the
plate circuit just as in an ordinary amplifier. Because of this
amplification, this type of detector is more sensitive than the C
bias type.
Such a tube detects on its grid-current curve, and so the grid
voltage must be such that the operating point is on a curved
part of the grid-voltage—grid-current curve, and because it amplifies
on its plate-current curve the plate voltage must be fixed so that
the operating point is on a straight part of the plate-current curve.
The grid current plotted against grid voltage of a 201—-A type
tube is given in Fig. 208. It will be noted that even though the
grid is negative a certain amount of grid current flows. This is due
to the few electrons which leave the filament with sufficient velocity
GRID LEAK AND CONDENSER DETECTOR 349
to get to the plate even through the negative retarding force of the
grid. This grid current must flow through the grid leak, usually
of the order of from 1 to 10 megohms. The voltage drop across
this resistance is such the the grid end is negative with respect to
the filament even though the “ grid return ”’ is connected to the
5
2)
Qa
—£ 4
<
fo) UX-201 A Tube
i) E;=5.0 Volts
= Ep=22 Volts
= 3 Grid Leak Resistance
=12 Megohms
Grid Return to F+
Rectified
Grid Current
Operating Grid N (ENE
Potential E> \ \ 1
—_|
WE Wwe]
ere eee ee ee Bee
0.4 —0.2 (0) +0.2 +0.4 +0.6
Grid Voltage, Eg
—0.6
Fig. 208.—Rectification on the grid voltage-grid current curve.
positive end of the filament battery. The value of the grid leak
fixes the point on the grid-current curve at which the input signals
operate. In Fig. 208 the operating point is about 0.3 volt negative.
A higher grid leak has a greater voltage drop across it and so the
operating point is somewhat more negative, but the entire range of
erid-leak values that are used does not vary the operating point
more than 1 or 2 volts.
350 DETECTION
278. Effect of grid leak and condenser values.—Changes in
grid leak value produce no other change in the detector action
than is produced by changing the operating point. This is the
entire purpose of the grid leak. The purpose of the grid condenser
is to by-pass the high resistance so far as r.-f. currents are con-
cerned so that the greatest possible r.-f. voltage may be built up
across the grid-filament input of the tube. If the condenser is too
small, there will be an appreciable r.-f. voltage loss in it, and the
tube will not get all possible of the input signal. If the grid con-
denser is too large, the audio-frequency voltages built up across
the grid leak will be by-passed. The grid condenser with modern
types of tubes should never be greater than 0.00025 mfd. and a
value of 0.0001 may be used satisfactorily. Smaller condensers
than this produce some loss in r.-f. voltage, and result in de-
creased sensitivity.
279. Detector action——When an input signal is applied to the
tube the grid voltage changes in accordance with the incoming
signal just as it does in the case of an amplifier or a plate current
detector. These changes in grid voltage produce a change in grid
current in accordance with movements up and down on the curve
of Fig. 208. Because of the curvature the grid current increases
more when the grid is positive than it decreases during the negative
half-cycles of input voltage. The result is a net change in grid
current, in this case an increase. This increase in grid current pro-
duces an increased voltage drop across the grid leak and a greater
negative voltage in the grid. This increase in bias causes a decrease
in plate current. It will be remembered that an input signal
caused an increase in plate current in the plate circuit detector.
These audio grid current changes produce corresponding plate
current changes whence they are passed on to the audio amplifier.
In practice, then, modulated r.f. voltages are put on the input
to a detector; within this detector the modulations are separated
from the carrier that brought them to the receiver; and finally
these modulations in the form of a.f. frequencies are applied to a
plate-circuit load—usually the input to an a.f. amplifier.
How this separation of carrier from modulation takes place, and
how much a.f. one gets out of a given detector with a given input
DETECTOR ACTION 351
modulated at a given percentage may be determined experimen-
tally as follows. We may fix upon some grid bias voltage, e.g.
in the case of a UY 227 about 18 volts with a plate voltage of
180. Then we can put on the input to this tube, operating as an
overbiased amplifier, or detector, various alternating voltages
which need not be at radio frequencies. These input voltages
Ip Milliamperes
0 50 100 150 200
E, Volts
Fic. 209.—Plate current curves as controlled by plate voltage and input carrier
voltage applied to grid.
cause some change in plate current which may be read with a fairly
good milliammeter. All that is needed then to experimentally
determine the detection characteristic of a C-bias detector is a source
of known alternating voltages and a good milliammeter.
Such a series of curves are shown in Fig. 209. For example with
100 volts on the plate, an input alternating voltage of 12 produces
852 DETECTION
a plate current of 1.5 milliamperes. Across such a family of curves
a load line (see Section 186) can be drawn for any load resistance,
in this case 200,000 ohms. This is about the highest resistance
load that can be used because of various capacities which will
shunt it and reduce its impedance at the higher audio frequencies.
The rectified output voltage of the detector, e.g., the a.f. voltage
applied across the 200,000 ohm resistor and hence to the input
of the a.f. amplifier, can be obtained from such a curve. For
example with an input of 12 volts (H = 12) the rectified voltage
may be found by noting the intersections of the load line with the
E = 0 line and with the H = 12 line. Thus the rectified voltage
is the difference between #, = 156 (intersection with H = 0) and
66 (intersection with H = 12) or 90 volts. Taking several of such
voltages a curve like that in Fig. 210 can be plotted.
Now this curve gives not only the rectified voltage due various
values of carrier voltage but by knowing how strongly this carrier
is modulated, the actual a.f. voltages applied across the load
resistance may be ascertained. For example, if the plate voltage
is 300 and H, = 27 volts, suppose a carrier voltage of 12 is modu-
lated 33 per cent. The carrier voltage will then vary between
, 12 — 12 X 33 per cent and
12 + 12 X 33 per cent or
between 8 and 16 volts.
These values of carrier volt-
age represent rectified volt-
ages of 47and 124 and because
the carrier swings as far up as
it does down from its unmod-
ulated value of 12, the audio
voltage produced by these
hae: . 124—47
variations is ae or 38.5
V Rectified
0 5 10 15 volts.
an E Carrier at 280. Power detection.—
4 a 1 7 ‘ ” . .
aay tlate:rectifications pow: Because of distortion and
detector characteristic. . : - E
hum arising in audio ampli-
fiers some designers believe it advisable to do away with at least
one stage of audio amplification. During 1929 many receivers were
POWER DETECTION 353
developed which had only one stage of audio amplification and the
expression arose of “ power detection,” meaning that the detector
put out enough a.-f. voltage to “load up ” the final power tube in
the receiver. How is this done? What is a power detector?
A power detector is simply a detector that will not overload
when very large r.-f. input voltages are applied to it and which will
turn out considerable power in its output. Here again there are
two kinds of detectors, the grid leak and condenser and the C bias.
The C bias is the more common type; it is simply a highly biased
tube. Even a power detector can distort; on Fig. 210 if the lines
curve appreciably, equal carrier voltage swings will not produce
equal a.c. voltage swings.
281. Distortion from square law detector.—It is an advantage
from the standpoint of efficient use of transmitting power to keep the
modulation at the transmitter as high as possible. It is an unfor-
nate fact, however, that with square law detectors the distortion
appearing in the output of the detector—due to second harmonics
generated in the tube—increase as the square of the modulation.
They are proportional to M?/4 where M is the percentage modu-
lation. Thus when an r.-f. wave is modulated 100 per cent there
are 25 per cent second harmonies present in the detector output.
With linear detectors high percentages of modulation can be
used without this second harmonic distortion, and it is probable
that all good broadcasting will use high modulation as more
receivers employ linear detection.
The advantages of power detection are: (1) possibility of linear
detection and consequent decrease in distortion; (2) ability to do
away with one stage of audio amplification and the distortion that
occurs in it; (3) shifting amplification from audio to radio decreases
noise both from hum and from tube noise itself; and (4) decreased
expense due to decreased filtering problem and decreased cost of
audio system.
A power detector can be arranged to overload before the final
power tube, or any other tube, in the receiver. Since the output
of a C bias detector decreases when its grid is forced positive, the
output from such a detector actually falls off with increase in input
beyond the overloading point. Thus it is possible to make the dis-
tortion from such a receiver actually less for a strong signal than
354
A. F. Ouput— Peak Volts
15
-05
DETECTION
A. F. Output— Peak Volts
Across R =50,000 Ohms.
(Resistance)
For 0.20 Modulation
VS R. F. Carrier
f= 8.02
Rg=1.0 Megohm R, = 10,830
7=-00025 MED
Ee
=+.5 Ef(Except C—327 E.=0) E,=24
Ef= =Rated Bitament Volts
CX—301A
Peak Volts Ps
4=30.3 |
R, =91,000
E,=69
Grid Leak Detection X—-112A
2
-050 -100 .150 .200 -250 .300
R. F. Carrier Peak Volts
Fra. 211.—Comparison of various tubes as grid leak detectors.
-350
DISTORTION FROM SQUARE LAW DETECTOR 355
for a weak one. In Fig. 213 will be found a curve showing how the
percentage distortion from the Crosley Jewelbox receiver (1929)
30
25
20
15
10
Ma.Rectified Plate Current
5 10 Ts 20 25 30
R.F. Input Volts
Fre. 212.—Comparison of grid leak and C bias detection.
30
28
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sue :
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2 ==--7-~~ | 1.0 Watt
% 40.20. 30 40 #50 60 70 #80 90. 100
Audio Volts Output
Fic. 213.—Overload characteristic of 1929 receiver. Increasing the input to
the amplifier to the detector-amplifier beyond the overload point results in
a decreased output (A). In older sets (B) distortion increases as input level
is raised.
falls off with greater output voltage whereas that of the coliven-
tional grid leak and condenser detector (B) not only increases
rapidly with output voltage but begins to overload at a much lower
output.
CHAPTER XV
RECEIVING SYSTEMS
THERE are three types of receiving circuits: (1) The tuned
radio-frequency receiver, neutralized or not; (2) the superhetero-
dyne; and (8) the receiver, like the Sparton, in which the processes
of selection and amplification go on independently of each other.
282. The tuned radio-frequency set.—The tuned r.-f. set of as
late as 1928 suffered from various faults: poor selectivity at high
radio frequencies, and excessive selectivity and poor gain at low
radio frequencies. In 1929 manufacturers began to look seriously
upon these faults with the result that the overall amplification
curves began to flatten out and the selectivity curves at 1500 ke.
began to look more like those taken at 550 ke. Some such receivers
still suffer from excessive sharpness of tuning so that higher audio
frequencies are cut off. The fact that tuning a receiver by varia-
tions of inductance reverses the faults possessed by a receiver
tuned by capacity, is used by several manufacturers to bring up
the amplification at low radio frequencies to a level comparing with
the gain at the higher frequencies. Others have other tricks to
make the response curve flat from 550 to 1500 ke.
283. The superheterodyne.—The double detector or super-
heterodyne receiver operates on a very interesting principle.
Theoretically is seems to have almost every advantage that other
receivers possess and some in addition. Practically it suffers from
some faults.
It is more difficult to get high amplification at broadcast fre-
quencies than at frequencies of the order of 30-200 ke. without
resorting to several stages, neutralizing circuits, careful shielding,
etc. Much greater amplification per stage can be obtained at lower
356
THE PHENOMENON OF BEATS 357
frequencies. If, then, we can change an incoming frequency of say
1000 ke. to 100 or even 50 ke., we can get as much amplification in
two stages as are obtained from four stages at the original fre-
quency. This means simplification in apparatus, and because of
the lower frequencies the problems of stability and shielding are
also simplified. An additional advantage lies in the constant band
width passed by the amplifier, regardless of the frequency of the
incoming signal. How is such a frequency change performed?
284. The phenomenon of beats.—Suppose two loud speakers
are attached to two oscillators, one generating a 1000-cycle tone
and the other an 1800-cycle tone. When these two tones enter the
ear, the listener hears not only the two individual tones but an
800-cycle tone too. If one of the two original tones is modulated
at another frequency, say 50 cycles at a given per cent, this 800-
cycle tone will be so modulated. By turning this modulated 800-
cycle tone into a demodulator, the 50-cycle modulations can be
got back.
The two frequencies above are said to beat with each other and
the difference frequency is called the beat note. If the two oscil-
lators are adjusted so that they are at zero beat they have the same
frequency and no beat note will be heard. In addition to the two
beating frequencies and the difference of the two (the difference or
beat frequency) there is a third frequency generated. This is the
sum of the two beating frequencies. It is called the sum frequency.
Thus in the above case there are in the ear 1800-, 1000-, 800- and
2800-cycle tones.
Now suppose we are receiving a 1000-ke. signal and want to
turn it into a 100-ke. signal. All we need is a local oscillator which
turns out either 1100-ke. or 900-ke. signals. We turn these two
signals into a mixing tube where the difference or beat frequency
is generated, put its output through a filter which cuts off every-
thing but the 100-ke. signal (which is now modulated at the same
modulations as the incoming 1000-ke. signal), and amplify it in an
intermediate-frequency amplifier. After sufficient amplification
has been attained the 100-ke. signal is put through a demodulator
and the original microphone modulaticns secured. These fre-
quencies can be put through an audio-frequency amplifier of con-
308 RECEIVING SYSTEMS
ventional design and the output finally put into a loud speaker.
The system is shown in Fig. 214.
285. Superheterodyne design.—Some superheterodynes use a
stage or two of radio-frequency amplification ahead of the fre-
quency changing system; others do not. Some have sufficient
amplification so that the input voltages are taken from a small
loop; others require an antenna of conventional form and size.
Some have one, two, or three stages of intermediate amplification.
In some the function of oscillation and frequency mixing goes on
in the same tube; in others these functions are separate. Some
systems amplify the sum of the two beating frequencies; most of
them utilize the difference frequency. Some have high and some
2nd A.F.
Detector Amplifier LS.
Viale.
Amplifier
Oscillator
750 or 850 K.C,
Fra. 214.—Symbolic diagram of superheterodyne.
low intermediate frequencies. Some use C bias detection and some
use grid leak and condenser detection. Some use air core and some
iron core intermediate transformers. And so on. Every designer
has his own ideas of which is best, or which avoids patents held by
some other designer. In one receiver of this type the beat fre-
quency was secured from mixing a locally generated second har-
monic of the incoming frequency with the signal frequency.
In all of the double detector receivers, the signals must first be
received and usually tuned—whether amplified or not—at the fre-
quency of the incoming signals. Then the mixing with the local
oscillator goes on, either in a separate tube—the first detector—or
in the oscillator tube, and then the unwanted products consisting
of the beating frequencies, ete., are filtered out. There need be
no more than two dials, one tuning the input circuit to the incoming
RADIOLA 60 SERIES 359
signal and the other tuning the oscillator until it is within the
required intermediate frequency of the incoming signals. This
produces the beat frequency which is passed through the inter-
mediate amplifier.
Modern design has eliminated one of the dials, so that both the
tuning process and the adjusting of the oscillator to produce the
required difference frequency are controlled by the same dial.
286. Radiola 60 series.—The curves given here are among the
few that have been presented in the radio literature on receivers of
this type. The data are representative of the Radiola 60 series
and were presented in a paper before the Institute of Radio
Engineers in March, 1928, by G. K. Beers and W. L. Carlson, of
the Westinghouse and General Electric Companies respectively.
This receiver series is composed of a tuned broadcast-frequency
amplifier, and a detector similar to a receiver of the T.R.F. type,
followed by an intermediate amplifier and second detector which
drives the single stage of audio amplification. The intermediate
frequency is 180 ke., keeping out of the second detector and the
loud speaker tube hiss and other extraneous and unwanted noises
which frequently pass through a lower-frequency amplifier which
acts to them as any radio amplifier would.
The radio-frequency amplifier differs from other amplifiers in
that the primary of the coupling transformer is larger than the
secondary. Although other transformers have small primaries
whose natural frequency is higher than any frequency that is to
be received, these transformer primaries are so large that they
resonate to a natural frequency lower than any frequency to be
received. At the lower broadcast frequencies, where most receivers
are insensitive, this amplifier has somewhat greater gain and
better fidelity characteristics. Where the old and small primary
increased the 2ffective secondary resistance at the higher fre-
quencies and thereby made tuning broader there, the new primary
makes the tuning broader at the lower frequencies and pulls up the
the amplification somewhat there, at the same time reducing it
somewhat at the higher frequencies. The result is a flatter amplifi-
cation curve as shown in Fig. 215. This scheme has been developed
by the Hazeltine laboratories for coupling the receiver to the
360 RECEIVING SYSTEMS
antenna. Better response at low radio frequencies is thereby
obtained.
Because of the higher inductance of the primary it has a capac-
ity reactance at all frequencies in the broadcast band, and being in
the plate circuit causes the tube to “ degenerate,” that is, to reduce
the input voltage because of the grid to plate capacity Just as an
inductive plate load causes the tube to regenerate. (See Section
258.) A small condenser acting as a neutralizing condenser reduces
the degenerating feed-back voltage and maintains the voltage gain.
The plate and grid coils of the intermediate amplifier trans-
former are both tuned and so coupled that the response curve is
broadened out, thereby providing a better transmission character-
Amplification (Voltage)
600 700 800 900 1000 1100 1200 1300 1400 1500
Frequency in K.C,
Fig. 215.—Amplification-frequency characteristic of r.-f. amplifier in
Radiola 60—gain per stage
istic by not suppressing side-bands as badly as many of the
amplifiers of the past. The overall response curve of the interme-
diate frequency amplifier is given in Fig. 216.
The second detector characteristic is illustrated in Fig. 210.
It is a UY-227 with 180 volts on the plate and 25 volts bias on the
grid. It is a power detector of the C bias type.
287. ‘‘ Repeat points.’’—Many superheterodyne receivers suffer
from ‘‘ repeat points,” or the fault of getting a station at two set-
tings of the oscillator dial, or getting two signals with a given setting
of the oscillator dial.
Let us suppose the beat frequency is to be 50 ke. At 550 ke.
the frequency of the oscillator can be either 600 ke. or 500 ke. to
provide this beat frequency of 50 ke. At 1500 ke. the oscillator
can be set at either 1550 or 1450 ke. This means that the range in
frequency of the oscillator must be from 500 to 1450 ke. or from
“REPEAT POINTS”. 361
600 to 1550 ke. Let us receive a station operating on 550 ke. If
the oscillator frequency range is from 500 to 1450 ke. , the 550-ke.
station can be heterodyned at two oscillator dial points, either at
500 or 600 ke. The 1500-ke. station can be heterodyned at only
one point, namely, 1450 ke., because it does not tune as high at
1550 ke.
Percent Responce
diola 6
150 160 170 180 190 200
KC
Fig. 216.—Response of intermediate frequency amplifier of Radiola 60.
On the other hand, if the oscillator range is from 600 to 1550 ke.,
the 550-ke. station can be heterodyned at only one oscillator dial
setting corresponding to 600 ke. because it does not tune as low as
500 ke., but the 1500-ke. station can be heard at two points, either
1450 or 1550 ke.
No matter which range the oscillator covers, stations operating
anywhere between 550 and 1500 ke. can be heard at two points;
362 RECEIVING SYSTEMS
thus an 800-ke. station can be heterodyned by setting the oscillator
at either 750 or 850 ke.
Now let us suppose the oscillator frequency is set at 750 ke.
If there are two stations, operating at 700 and 800 ke., equally
powerful at the input to the receiver, both will provide the required
50-ke. intermediate frequency and both will be received.
In two-dial receivers of this type, some discrimination against
unwanted stations is had by tuning the input circuit to the desired
station. Thus if the oscillator is tuned to 750 ke. and the input
circuit to 700 ke., the 800-ke. station would be tuned out and
would not affect the receiver.
But in single-control receivers both the input and the oscillator
are tuned at the same time. If, then, the control dial is set at some
point, two stations differing in frequency by twice the intermediate
frequency (in this case by 100 ke.) and equidistant (in frequency)
from the oscillator frequency will be received with equal strength.
This is because it heterodynes both of them by the required 50-ke.
frequency.
There is another cause for repeat points. If the oscillator gen-
erates harmonics, they may heterodyne a higher frequency incom-
ing signal according to the following reason. Suppose the oscillator
is set at 600 ke. but generates a strong second harmonic, 1200 ke.
This harmonic will provide the proper 50-ke. beat note with either
a 1150- or a 1250-ke. station at the same time the fundamental
heterodynes a 550 or 650 station. Such harmonics can be reduced
by making the oscillations feeble by reducing the plate voltage or
by placing a large resistance in the plate circuit to make its charac-
teristic straight, or by giving the grid the proper bias.
Let us suppose now that the beat frequency is 600 ke. The
oscillator frequency range must be from 550 plus 600, or 1150 ke.,
to 1500 plus 600 or 2100 ke., and there will be no repeat points due
to heterodyning two incoming signals to give the required 600-ke.
beat frequency.
The use of a radio-frequency amplifier ahead of the frequency
system will provide considerable discrimination against unwanted
stations. It must be tuned to the required station at the same
time the oscillator is tuned so that it gives the required beat fré-
FREQUENCY CHANGERS 363
quency. If the radio-frequency amplifier provides a 40-DB dis-
crimination against a station 20 ke. away from the desired station,
it will keep out of the remainder of the receiver signals from
undesired stations even though their frequency difference is correct
to pass through the intermediate amplifier.
288. Choice of the intermediate frequency.—If the inter-
mediate frequency is high there will be some difficulty in getting
a high voltage amplification. If the frequency is low, the amplifier
will probably transmit high audio-frequency noises originating
within the radio circuit. The high-frequency amplifier will suffer
from stray couplings between stages, and therefore will be difficult
to control. The low-frequency amplifier will make impossible com-
plete isolation of the oscillator and input circuits which are tuned
to within a few kilocycles of each other.
In the Radiola 60 series, the intermediate frequency is chosen
at 180 ke. as a compromise betwen these two difficulties.
289. Selectivity of superheterodynes.—In addition to the
selectivity provided by the r.-f. amplifier, there is some selectivity
added in the intermediate-frequency amplifier. Let us suppose
the desired signal gives the required 180-ke. beat frequency which,
according to Fig. 216, gives 100 per cent response. An unwanted
station operating at 20 ke. off the desired frequency is also hetero-
dyned by the oscillator, but instead of producing a beat frequency
of 180 ke. it produces either 160 or 200 ke. According to Fig. 216
these signals would be reduced to about 1.0 per cent of the desired
signal. This is an additional discrimination of 40 DB which added
to the 40 DB secured in the r.-f. amplifier provides a total selec-
tivity of 80 DB.
Problem 1-15. If the intermediate frequency of the Radiola 60 is 180 ke.,
what must be the range of the oscillator frequency to produce this beat fre-
quency at broadcast frequencies?
290. Frequency changers.—Any tuned r.-f. receiver can be con-
verted into a superheterodyne by the addition of an oscillator and
a mixing tube. In such a system the r.-f. amplifier is used as the
intermediate-frequency amplifier, and: is set to give maximum
amplification at some fixed frequency within the broadcasting band.
Then the oscillator beats with the incoming signals so that this
364 RECEIVING SYSTEMS
frequency is generated and the signals are finally detected in normal
manner.
291. The “autodyne.”—It is possible to do away with one tube
by combining the functions of oscillator and mixing tube, or first
detector. It is only necessary to couple the input circuit, the
antenna-ground system for example, to the oscillator which acts as
detector. The latter is tuned so that its frequency differs from
the incoming frequency by the number of kilocycles to which the
intermediate amplifier is tuned. Such a “super” is called an
autodyne because the signal is automatically heterodyned in the
local oscillator, or first detector, instead of requiring a third or
mixing circuit.
292. ‘‘ Short-wave” receivers.—The majority of the traffic
carried on in the higher frequency bands, from 1500 to 15,000 ke.,
is in code and it is not necessary to transmit a very wide range of
frequencies to convey good signals. In speech the frequency band
must be at least 5000 cycles wide, in transmitting music the band
should be 5000 each side of the carrier frequency or a band of
10,000 cycles. For code transmission the width of band required
depends upon the number of signals to be transmitted per second
and upon the nature of the signal.
The usual short-wave receiver for code reception consists of an
“autodyne”’ detector, that is, an oscillating detector which is
detuned from the incoming signal by about 1000 cycles. The plate
circuit has a low impedance to both the locally generated and the
incoming frequency but a high impedance to the 1000-cycle note
which is amplified by an ordinary audio amplifier and then passed
into a pair of head phones. Frequently a screen-grid tube is used
between the detector and the antenna, to provide somewhat greater
amplification, to prevent interaction between antenna and detec-
tor, and to prevent oscillations in the detector circuit from getting
into the antenna and being radiated from it.
Of course it is possible to make the beat frequency between the
detector oscillation and the incoming signal some intermediate fre-
quency and to pass it into an intermediate-frequency amplifier.
Many of the short-wave “adapters” consist of such apparatus.
The intermediate-frequency amplifier can be the usual radio-fre-
SHORT-WAVE RECEIVER CIRCUITS 365
quency amplifier from a broadcast receiver. Amplification takes
place in such a set at the wavelength to which the broadcast band
receiver is tuned, usually about 550 ke.
Other adapters are merely oscillating detectors which are
plugged into the detector socket of a broadcast receiver. The
tube gets its filament and plate voltage from this socket, and
passes its 1000-cycle beat note into the audio amplifier of the
broadcast receiver.
293. Short-wave receiver circuits.—The short-wave receiver
detector can be made to oscillate in any of the standard regenera-
tive circuits. Since the detector is to work at very high frequencies
Ll Ll
Amplifier
To
Amplifier
Le we ae ee om
b
a
fi
oH
A B+45V A B+45V
(A) (B)
Fig. 217.—Two methods of connecting feed-back circuit in short wave
receiver. A preferred circuit is in Fig. 221.
where any stray capacities or shunt paths become of great impor-
tance, several circuits have come through the development period
to be almost standard, combining ease of adjustment and stability
of operation. In Fig. 217 are shown two common circuits. They
differ in the manner of securing regeneration. They employ a
fixed tickler and a variable capacity adjustment. The variable
tickler has been abandoned in short-wave circuits because of
variations in tuning when its position is varied with respect to the
secondary winding. In Fig. 217A the r.-f. currents are prevented
from going into the a.-f. amplifier and made to go through the
regeneration system by means of the choke coil, RFC. If the
choke is poor, or if good coupling between tickler and secondary
366 RECEIVING SYSTEMS
coils cannot be obtained so that regeneration is not under good
control, the series arrangement in 217B may be used. Here all of
the plate circuit currents, a.-f. and r.-f., go through the choke coil.
Such a circuit is especially useful if the choke coil does not have a
high impedance over the band to be tuned to.
Since the choke coil is directly across the tickler coil and regen-
eration condenser shown in Fig. 217A (the B battery is at ground
potential and so is the filament), its low impedance at some fre-
quency may prevent oscillations. This choke coil because of its
inductance and capacity will have a resonant frequency, and some
harmonic of it may appear within the band to which the receiver
may be tuned, or the choke may tune within the band by means
of some series capacity. Under these conditions the operation of
the receiver will be erratic.
L-—ww
il
Fig. 218.—Two methods of coupling antenna to input of receiver.
294. Coupling the short-wave receiver to the antenna.—An
oscillating tube may be connected directly to the antenna either
through a coupling coil or through a small condenser (Fig. 218) or
through a coupling tube, usually a screen-grid tube which
may or may not be arranged with its input circuit tuned. If the
input circuit is tuned, care must be taken to prevent feed-back
between its output and this antenna-ground input inductance.
USE OF SCREEN-GRID TUBE AT SHORT WAVES 367
Otherwise the tube and circuit will oscillate, and stable reception
will not be possible.
If the antenna is connected either through an inductance or a
condenser, some harmonic of the antenna may fall within the band
over which the receiver is to be tuned. It will then absorb energy
from the oscillating detector and it will be difficult to make the
detector regenerate properly. Such difficulties result in “dead
spots ” and will require a looser coupling to the antenna to prevent
interference with the tuning of the receiver. The use of a coupling
tube will eliminate dead spots. Its input should be tuned.
295. Use of screen-grid tube at short waves.—Figure 219, taken
from the Hammarlund Short Wave Manual, 1929, is the equivalent
circuit of the screen-grid
tube when used between
two impedances, Z; in-
put and Z, output. The
grid-plate capacity is
represented as C,, and
the plate resistance of
the tube in series with
the. input voltage mul-
tiplied by the w of the Fic. 219.—Equivalent circuit of screen grid
tube is shunted across tube and input and output impedances.
the circuit. The pro-
portion of voltage usefully employed across Z, may be found
by comparing its impedance to the total impedance of the
circuit DEFG. If this impedance is 900,000 ohms and if Z is
100,000 ohms, the voltage across Z will be roughly one-ninth of
the voltage developed in the tube, u#,, and the amplification will
be one-ninth of yu, or for the average screen-grid tube about 30.
The same result may be calculated by multiplying the mutual con-
ductance of the tube by the output load, 300 X 10~® x 100,000.
This is because the load, Z, is small compared with the R, of
the tube.
At high frequencies the reactance of C,, becomes fairly small
and so energy fed into Z from the generator, wE, may get back
to the input circuit, Z;. This results in regeneration and if this
368 RECEIVING SYSTEMS
feed-back voltage is equal to the original voltage across AH due
the incoming signal, continuous oscillation will probably result.
Such a condition is undesirable. If the amplification in the tube
is 30, 30 times as much voltage exists across Z, as across Z;, and
if the reactance of C,,, is low enough so that the voltage fed back
to Z; is one-thirtieth of that across Zo, oscillations result. This
will be true when the frequency gets high enough so that the
reactance of C,, becomes sufficiently low. If Z; and Z, are each
equal to about 100,000 ohms, the limiting frequency that can be
received with stability is
such that the reactance of
C,, is 80 times this value
or 3,000,000 ohms. Such
a frequency is 1770 ke.
Frequencies higher than
this will cause trouble.
One of two things must
: Bottom View
‘ of Pugtor take place to prevent re-
A- A+ B+45V B+135V Detector Socket ? Ms
Fic. 220.—A screen grid blocking tube be- SEES GUESS a0 A,
tween antenna and oscillating detector; all must be decreased. De-
arranged to be plugged into a broadcast or creasing Z; decreases the
other receiver. proportion of voltage that
is fed back, and decreas-
ing Z, decreases the amplification. If the antenna is connected as
in Fig. 220 through a 5000- or 10,000-ohm resistor, trouble from
oscillation will not occur. If, on the other hand, the input is
tuned (Fig. 221) oscillation may result. The input shown in Fig.
220 is prone to pick up local interference. Fig. 221 is better.
It becomes a question of extracting the maximum voltage
amplification possible from the tube, or from getting less amplifica-
tion and building up the signal strength by boosting the input
voltage to the tube. Tuning the input and keeping the output
somewhat low in impedance may result in greater overall amplifica-
tion than is possible by a low input and high gain. Coupling the
screen-grid tube to the detector through rather loose coupling will
lower the effective resistance in the plate circuit of this tube and
decrease the amplification but increase the stability, and permit a
LONG-WAVE RECEIVERS 369
greater input impedance to be built up in the screen-grid
stage.
Adjusting the coupling condenser C, limits the amplification to
the value that is stable because the impedance in the plate circuit
of the screen-grid tube is low enough to prevent oscillation.
296. Long-wave receivers.—The usual long-wave receiver uses
an autodyne detector too, but because of the detuning effect,
which may be serious at low frequencies, a better scheme is to use
a separate oscillator and to beat it with the incoming signal.
When one listens to long-wave stations for the first time on
such simple receivers, he is struck by the fact that many stations
are heard at the same time. The reason is as follows. These
2 o 6 %OB+4
A A+ B+135V
Fria. 221.—A short-wave detector circuit preceded by a tuned amplifier.
stations operate on the frequency band from 10,000 to 20,000
meters or from 15 to 30 ke. When a receiver is tuned to a 20-ke.
station, it is detuned only 1000 cycles—an audible amount—from
a 21-ke. station, only 2000 cycles from a 22-ke. station, and so on.
If the operator tunes to the 10,000-meter—30-ke.—station he is
only detuned by 15 ke., an audible amount, from the station
allotted the channel at the extreme other end of the band.
In commercial receiving stations the signals are tuned and
filtered so that a very narrow band is passed, about 200 cycles.
In this manner stations can be separated. All of the long-wave
stations in the world are in this limited band, and of course any
station can be heard in any other part of the world,
370 RECEIVING SYSTEMS
The coils frequently used in receiving high-power long-wave
stations are the commonly known ‘“ honeycomb ”’ coils and are
highly concentrated multi-layer inductances. A table showing the
wavelengths to be received with certain sizes of coils is given
herewith.
TABLE I
Induc- Dis- _ Wavelength Range, Meters
Natural :
Number | tance, at Waves tributed
of 800 Cycles, Tete Capacity
Turns | in Milli- ate in 0.0005-mfd. 0.001-mfd.
henrys Muifd. Condenser Condenser
2D 039 65 30 120 to 245 120 to 355
35 .0717 92 33 160 to 335] 160to 480
50 .149 128 31 220 to 485] 220to 690
75 325 172 26 340 to 715] 3840to 1,020
100 . 555 218 24 430 to 930 430 to 1,330
150 1.30 282 il/ 680 to 1,410} 680to 2,060
200 2a 358 16 900 to 1,880} 900 to 2,700
249 3.67 442 15 1,100 to 2,370 | 1,000 to 3,410
300 Deso 535 ily/ 1,400 to 2,870] 1,400 to 4,120
400 9.62 656 13 1,800 to 3,830] 1,800 to 5,500
500 15.5 836 13 2,300 to 4,870 | 2,300 to 2,000
600 21,6 1045 14 2,800 to 5,700 | 2,800 to 8,200
750 34.2 1300 14 3,500 to 7,200 | 3,500 to 10,400
1000 61 1700 13 4,700 to 9,600 | 4,700 to 13,800
1250 102.5 2010 iil 6,000 to 12,500 | 6,000 to 18,000
1500 155 2710 13 7,500 to 15,400 | 7,500 to 22,100
Experiment 1-15. Connect three honeycomb coils as shown in Fig. 222
making S a 1500-turn coil, P about 1000-turn, and 7 about 750-turn. The
tuning condenser can be either 500 or 1000 mmfd., preferably the latter.
Connect an antenna as long as possible to the 750-turn coil and listen in the
plate circuit. It may be necessary to reverse the connections to the coil 7 in
order to make the tube oscillate. It should be possible to hear many long-wave
stations transmitting traffic to foreign countries. The phenomenon of zero
beat is beautifully illustrated by such an experiment. Some of the long-wave
stations that should be heard are;
DETUNING LOSS IN AUTODYNES 371
New Brunswick, N. J....WII.......: 13,750 meters
New Brunswick, N. J....WRT....... 13,265 meters
BolmasyCahte se. oe.) IIDOh oo ee ce 12,500 meters
Rocky Point, N. Y...... W'ELXO aes 15,806 meters
RockysPoints NewYennene WOKE Te 16,465 meters
iodlay Rome, IN, We. 5 WSSAriee ee 16,120 meters
diueckerton Ni Jee WiCiea eer. 16,700 meters
eliveker tons Nese WC Gy aeerrss 15,900 meters
To AF. Det.
>
Amplifier
Osc.
' | (B)
+B
T
Fic. 222.—The beat note in a code receiver may be obtained in the “auto-
dyne”’ manner (A) or in heterodyne (B).
297. Detuning loss in autodynes.—Some loss in signal strength
is experienced in autodynes because the detector is actually detuned
from the incoming signal. At high frequencies where such a system
is frequently used, the detuning is not serious. Thus at 30 meters,
10,000 ke., a deviation of 100,000 cycles is only 1.0 per cent. At
the longer waves, however, detuning only 1000 cycles to get an
audible beat note represents an appreciable loss. Thus at 20,000
meters, 15 ke., a detuning of 1000 cycles represents a detuning of
over 6 per cent. The use of a separate oscillator, as in Fig. 222B,
will prevent this loss.
Problem 2—15. The following voltages were measured across a 600-turn
honeycomb coil when it was tuned to 40 ke. and the input frequency was
changed. How much loss in signal strength would be incurred if the coil and
condenser were used in an autodyne detector and detuned from 40 ke. so that
372 RECEIVING SYSTEMS
a 1000-cycle bear note was secured? Plot the curve of voltage against fre-
quency and note the loss from it.
TABLE II
Frequency Voltage Frequency Voltage
40.0 2.5
40.5 ib 3
41.0 .95
41.5 .05
298. Poor quality on long waves.—It is much more difficult to
transmit or receive high quality music or speech on the longer
waves. Suppose the transmitter is tuned to 10,000 meters, 30 ke.
The band transmitted must be 10 ke. wide or 30 per cent. At
broadcast frequencies, however, the band passed is only 10,000
cycles in a mean frequency of 1000 ke. or 1.0 per cent. This means
that very broad circuits must be used at intermediate or low ratio
frequencies, which in turn means poorly selective circuits.
299. ‘Band pass’ receivers.—The third type of receiver in com-
mon use is due to the desire to provide greater fidelity of reproduc-
tion and at the same time to increase the selectivity of the receiver.
The ideal response characteristic is a flat-topped curve with very
steep sides. Such a curve would provide good transmission within
the desired band, say 10,000 cycles, and nothing at all beyond
that band. The resonance curve of the ordinary tuned circuit, or
of several such circuits in cascade, has a narrow top, and more
gently sloping sides than is the ideal.
By the use of coupled circuits it is possible to approach the ideal.
Thus in Fig. 223 if both coils are individually tuned to the same
frequency, and then coupled together, the resultant response char-
acteristics may be a single narrow-topped curve, like that of a
single tuned circuit, or a flat-topped curve, or a curve of two more
or less widely separated peaks with a hollow between, depending
upon the degree to which the circuits are coupled.
“BAND PASS” RECEIVERS Biss
In Fig. 223 is given the result of coupling two such circuits
together with various degrees of coupling. It will be seen that too
close coupling gives the widely separated peaks, proper coupling
gives a comparatively flat-topped characteristic, and too loose
coupling gives a sharply tuned circuit, cutting side-bands as badly
as a single circuit, although its sides drop more precipitously than
Single
——
ut=.026 Volts
> (aT
¥=560 KC
0 10 20 30 40 50 60
KC
Fig. 223.—Experimental determination of effect of close coupling in band
pass filter at high frequencies.
a single circuit and thereby provide greater selectivity against
stations at some distance from the resonant frequency.
Superheterodynes now use such a “ band pass ”’ circuit in the
intermediate frequency amplifier. Here the frequency is fixed, and
one adjustment will do for all signals put into it. When the band
pass arrangement is used at broadcast frequencies, the width of
band passed may differ at each frequency to be received. If the
coupling is by inductance the band will be broad at the high fre-
374 RECEIVING SYSTEMS
quencies; if the coupling is capacitive, the curve will be broad at
low frequencies. Some combination may be arranged so that a
more or less uniform band is passed at all broadcast frequencies.
The ‘‘ band pass ”’ arrangement may be placed between tubes,
or several stages of it may be placed ahead of the amplifier, and
thus used to filter out the desired band before any amplificatien
takes place. In the latter case an untuned amplifier is necessary,
that is, a system which amplifies all frequencies more or less alike.
300. The Sparton receiver.—The first commercial receiver to
use the ‘‘ band pass” arrangement was the Sparton, which had a
band selector and an untuned amplifier. The band selector was
To
—> Amplifier
Fiq. 224.—Band pass circuit.
composed of several tuned circuits coupled together by means of
small inductances as shown in Fig. 224. The amplifier had a voltage
gain of about 5000 at 1500 ke. to about 18,000 at 550 ke. In
operation the band selector was adjusted so that it admitted and
transmitted the desired station’s signals. These were amplified by
the untuned amplifier, and rectified by the power detector and
thence transmitted to the power amplifier.
301. Experiments with band pass filters—The results of mak-
ing some laboratory measurements on such filters may be seen in
Fig. 225. These data were collected by Kendall Clough and pub-
lished in Radio Broadcast, December, 1928. They show the diffi-
culty in getting a uniform band, or uniform transmission loss or
gain, over a wide frequency band. The dotted curves are of a
single tuned circuit. The curves show gain in a single stage.
EXPERIMENTS WITH BAND PASS FILTERS 375
302. Measurements on radio receivers.—It is now possible to
make very comprehensive and thoroughly quantitative tests on
radio receivers, either as a whole or upon the component. parts.
Such tests consist in measuring what comes out of a receiver when
known voltages at known frequencies modulated at known per-
centages are placed on the antenna-ground binding posts of the
receiver.
The voltage on the antenna-ground posts is secured in several
ways. Some laboratories use an artificial or “dummy ”’ antenna
Voltage Amplification
12 7 12
11 a il 1l {>
10 — 10
9 Se 9
i. 8
7 oe 7
6 : 6
5 ; : 5
41 4
3 + 37
2 a 2
1 : 1
-20 -10 600 +10 +20 $o -10 1000 HO +20 So -10 1500 +10 +20 +30
KC KC KC
(A) (B) (C)
Fig. 225.—Quantitative data on band pass amplifier-tube and tuned coils.
The dotted curves are of a single coil properly tuned.
consisting of concentrated capacity and inductance and resistance
of such values that they simulate the antenna-ground system
ordinarily used. Such values used in some laboratories are: 200
microhenrys, capacity 200 mmfd., and resistance 25 ohms. Other
laboratories use a coupling coil into which the desired voltage is
induced, and others take the voltage drop across a known resist-
ance which is in series with the artificial antenna. If comparative
measurements and not absolute are required, the output from the
receiver when it is attached to an antenna of the usual type can be
employed. Of two receivers the one which gives more output from
a given station on a given antenna has the greater overall amplifica-
tion.
A erystal rectifier and meter will measure output.
376 RECEIVING SYSTEMS
The load into which the output of the receiver is measured is
usually a non-inductive resistance and the standard output is
taken at 50 milliwatts. Sometimes a loud speaker is placed across
the receiver output and the current into it and voltage across it
are measured with given input voltages and at various frequencies.
These volt-amperes plotted against input frequency or against
input modulating frequency give an indication of the overall volt-
age amplification as well as the overall fidelity characteristics of
the receiver.
Although it is important to measure the individual components
out of which the receiver or power equipment or transmitter is
made, the overall characteristic tells the story of exactly what the
apparatus as a whole will do.
303. Signal generator.—For receiver measurements some
means must be provided for furnishing known amounts of radio-
frequency voltages. Since the modern radio receiver has a very
high voltage and power amplification, these voltages must be very
small when it is desired to measure the overall characteristic or
performance. It is desirable to have a known voltage at least as
low as 1 microvolt, and anyone who has worked intimately with
radio-frequency voltages—at one million cycles for example, knows
how difficult it is to know when one has an e.m.f. of this order or
a current of one millionth of an ampere. The laboratory worker
must know how much current or voltage he has, and he must be
certain that his meter shows all the current or voltage, no more
and no less.
The circuit diagram in Fig. 226 is that of the General Radio
Signal Generator, a device which consists of a radio-frequency
oscillator, a means of measuring and controlling its ouput, and a
means of using any desired part of this output for purposes of
measuring receivers, either as a whole or in component parts.
In practice a fairly large current is measured, which flows
through a known resistance and provides a fairly large voltage.
Then an “ attenuator” or voltage decreasing system is provided
which has been calibrated, preferably in DB, so that any desired
known part of this fairly large voltage can be presented to some
outside circuit.
MODERN RECEIVERS 377
304. Modern receivers.—Some sensitivity and _ selectivity
curves of a modern receiver (Stromberg Carlson Model 641) will
Standard
Signal
Generator
‘= Ground
Ex
M
0.25 E05
Modulation
Control
Tos
~ Radio Control
+05
Radio Current
Meter
Modulation Voltmeter
Fra. 226.—Cireuit diagram and method of using signal generator.
be found in Figs. 227 and 228. They are characteristic of the
better sets made in 1928-1929. Such receivers are more sensitive
than is necessary and probably would not be made to deliver a
378 RECEIVING SYSTEMS
normal output on such low field strengths were it not for com-
petition and the lure of “ Dx.”
1400
— CM'W 0S) 3Nd3yn9 psepuejs 404 yNdu| s}}OA0I9 |
fy
o
5
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ai
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10 20
Higher Frequencies
Kilocycles off Resonance
Selectivi
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Lower Frequencies
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1000
tere adueuosay je jndu|
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Fig. 228.—Selectivity curve of 1929 receiver,
A compared to 1928, B
Sereen-grid tubes make possible receivers with fewer tubes,
greater sensitivity, greater selectivity, and perhaps better fidel-
ity. Two screen-grid tubes should deliver a voltage amplification
at average broadcast frequencies of 2500 and if there is a voltage
APPARENT AND REAL SELECTIVITY 379
gain of 5 in the antenna or another stage of amplification tremen-
dous voltage amplification becomes possible. Just what can be
done with such amplification is a question. In most cities the noise
level is so high that a receiver probably can never utilize its full
amplification.
The possibility of enormous amplification preceding the detec-
tor eliminates considerable distortion occurring in weak signal
detectors by working them at higher levels where linear detection
takes place, removes some hum from the output, and considerable
audio-frequency regeneration. At the same time the screen-grid
tubes make possible a more selective receiver (their resistance is
very high and can be shunted across tuned circuits without increas-
ing the resistance of the latter so much as when 12,000-ohm tubes
are employed) and such selectivity may imply that some high
audio tones are cut out. To get around this difficulty manufac-
turers are experimenting with band pass tuning which, theo-
retically at least, presents a flat-dopped, steep-sided resonance
curve.
305. Apparent and real selectivity.—The fact that the sensi-
tivity and selectivity of a receiver are interrelated has been men-
tioned. Even in those few receivers in which the two functions
are separated, there is an apparent relation between these two
important factors. The more sensitive a receiver is the less
selective it seems to be. This is because the more sensitive it is,
the more it amplifies weak signals. On a less sensitive receiver,
signals on a frequency channel near the one to which the set is
tuned may be so weak they do not bother the desired program.
If the sensitivity of the receiver is increased, the desired signal
increases in level but so does the unwanted signal. Both may
increase in the same ratio, so that the real or measurable selec-
tivity remains the same, but the unwanted station now begins to
affect the listener’s ear. When the desired station is quiet, during
announcements for example, the unwanted station comes in.
The answer is to decrease the sensitivity of the receiver so that
the unwanted signal is again pressed below the level (A in Fig. 229)
where it is bothersome, and if more volume is required to increase
the audio-frequency amplification. Thus a receiver should have
380 RECEIVING SYSTEMS
a sensistivity control and a volume control. The first operates on
the radio-frequency amplifier, the latter may operate on either the
radio- or the audio-frequency am-
plifier. In Fig. 229 the vertical
rn lines represent broadcasting signals.
In B, three of them are being re-
ceived, the 1000 ke. the loudest.
B In A only one signal, at 1000 ke. is
500
ray
i=]
Oo
entering the receiver.
10 306. Importance of shape of con-
: = denser plates.—A receiver can be
ig : calibrated so that each degree of
Field Strength Microvolts per Meter
its dial may represent one channel
Wy A . -
1980 990 1000 1010 1020 Width, 10 ke., or it may be cali-
Frequency in Kilocycles brated in wavelengths so that turn-
Fria. 229.—Illustrating real and ing the dial a certain number of
apparent selectivity. degrees turns through a certain
number of meters. Or a combina-
tion of “straight frequency line” and “straight wavelength
line”? tuning may be effected. With a semicircular plate con-
denser a greater number of frequency channels, each 10 ke. wide,
will be passed over by a given number of dial degrees at the
high frequencies than at the low, and the effect upon the opera-
tor is an apparent decrease in selectivity of his receiver at short
wavelengths.
307. Automatic tuning.—Several schemes have been proposed
for automatically tuning a receiver; some of them have been
patented and utilized in commercial models of receivers. Such a
receiver has a switch or a button, or a lever, on which is marked
the station desired. It is only necessary to operate the device to
tune the receiver to the desired frequency. Pressing another button
tunes in another station.
Such systems can be operated with electrical relays, one relay
setting the circuits for each frequency, or by mechanical devices,
levers, motors, ete.
308. Automatic volume control.Some advance has been made
toward automatic volume control for radio receivers. Anyone who
SHIELDED RECEIVERS 381
has operated a receiver at such a distance from the transmitter
that fading is pronounced will appreciate the need for such circuits
The use of such a volume control means that once the adjustment
for maximum sound output has been made, tuning the receiver
through local and distant stations will not produce blasts of dis-
torted sounds when passing the carrier wave of a strong station.
Provided the distant station puts down a sufficient field strength
near the receiver, it can be brought in with the same volume as a
local station. The difference in the majority of cases will be in the
noise that comes along with the station’s signals. That is, if the
distant station is much weaker than the local station, the sensi-
tivity of the receiver will be increased automatically by the control
system to the point where the distant station gives loud speaker
signals. In this sensitive condition the receiver will admit other
extraneous radio-frequency disturbances, such as static, power
leaks, ete.
One volume-control circuit is shown in Fig. 230 and is applied
to the Radiola 64. It is simply a tube placed in parallel with the
second detector of the receiver.
=
Audio Detector
When loud signals are received
this tube has its grid voltage
decreased and hence its plate
current increased. This in-
creased plate current produces
a voltage drop across a re-
sistance and is then applied to
the grids of the radio- and
Output
Volume Control
intermediate-frequency ampli- Tubes’ pag Cathode Plate
fier tubes thereby increasing Amplifier Tubes
the negative bias and reducing yg, 230.—Automatic volume control
the overall amplification. Some circuit.
manual adjustment is provided
so that the output level at which the volume control begins to
work is under control of the operator.
309. Shielded receivers.—The greater the voltage gain in a
radio-frequency amplifier, the more necessary it is to prevent any
energy from a tube near the detector from feeding back into a tube
382 RECEIVING SYSTEMS
near the antenna. Otherwise oscillation results and the receiver
is useless. One method is to neutralize the grid-plate capacity
within the tube. Another is to filter each plate and grid lead so
that the r.-f. currents stay where they belong and do not get
into the plate voltage supply system. Still another is
shielding.
Now placing a coil in a metal box seems like a simple trick to
keep the lines of force from that coil from getting mixed up with
lines of force from another coil. But the size of the metal box, its
material, the size of the coil, whether or not the metal box is
grounded, and to what, and whether it carries current or not—all
of these things enter into the problem of shielding.
If the lines of force from a coil are stopped by the shield, cur-
rents are set up in this metallic shield. These currents, flowing
through the resistance of the shield, represent a certain amount of
power lost. This power must be supplied from the tuned circuit,
and is thereby subtracted from the useful function of providing
signal strength to actuate the grid of some succeeding tube.
The subtraction of energy from the tuned circuit by the shield
has the same effect as though the resistance of the tuned circuit
were increased. In other words, if currents are induced in the
shield, the effective resistance of the coil is increased, its selectivity
factor goes down, the voltage gain due the coil and condenser is
decreased. It is also true that the inductance of the coil decreases
too, so that the value of effective resistance for the tuned circuit,
[?w*/r is greatly decreased. The nearer the coil is to the shield,
and the greater the resistance of the shield, the greater is the power
loss in it. If the shield is made up of sections which slide into each
other and these sections do not fit tightly, a joint of high resistance
may result. This means that not only is the shield ineffective but
that it consumes too much useful energy from the tuned circuit.
The shield must be large compared to the coil.
The shield should never be used to carry current. That is,
attaching a filament wire to one end and the battery to another is
bad practice. It is still worse practice to make it a part of any
tuned circuit. The shield should be made heavy and from the best
conductor economically possible, and all joints in it should be
LOUD SPEAKERS 383
carefully soldered. It should be connected at only one point to a
heavy conductor leading to the common ground of the set. Holes
in it for leading-in or -out wires should be small.
From the electrical standpoint, copper is better than brass or
aluminum; from the standpoint of weight and cost per pound,
copper is handicapped.
It is interesting to consider the circuit as a transformer and its
load. The resistance of the shield represents the load across the
transformer. The power in this load is taken from the generator
which is the tuned circuit. As a matter of fact any two wires
which couple any two circuits together may be considered as a
transformer, one acting as the primary and the other as the secon-
dary. The power wasted in the load on the secondary wire must
come from the generator or the circuit attached to the primary
wire and represents an increase in the resistance of the circuit
attached to this wire, and a decrease in its inductance.
310. Loud speakers.—The loud speaker is the final link in the
broadcasting system, and because of its position with respect to the
listener it is frequently blamed for much of the bad reproduction
that really originates somewhere else.
The task of the loud speaker is to translate into sound energy
the electrical energy in the power tube. It must do this as effec-
tively and faithfully as possible. It is useless to design and operate
a high-class amplifier with a poor loud speaker. The wide range of
tones coming from the amplifier is lost in the loud speaker and does
not get to the listener. Likewise, it is absurd to install a perfect
loud speaker in the hope that the fidelity of reproduction from an
antiquated or poorly engineered receiver will be bettered. ‘The full
benefit of a perfect loud speaker cannot be attained until the com-
plete chain of apparatus is perfect—amplifier, power tube, plate
voltage supply, loud speaker, etc.
Loud speakers in general are notoriously inefficient—the best
in common use is not over 30 per cent. Most of them are less than
5 per cent efficient. Some of them reproduce the whole audio range
of tones with almost equal fidelity. Others reproduce only a very
small section of the audio spectrum—such were the short-horned
loud speakers of a few years ago.
384 RECEIVING SYSTEMS
311. The horn type.—Any device which will move when a
modulated electric current flows in it will make a loud speaker.
Its movements are imparted to the air which in turn affect our ear
drums and our auditory nerves in a sensation we call sound. The
object is to effect as large a movement of air as possible with the
least possible electrical input, and to effect this efficient transfer of
electrical into sound energy over as wide a band of audio tones as
possible.
The horn type of speaker may use a thin steel or iron diaphragm
as the moving element or a non-magnetic diaphragm actuated by
a mechanical coupling system as in Fig. 231. The electric currents
are sent through the winding of a
Neca nearby electromagnet which has a
Coil certain amount of permanent mag-
| Armature netism in it. When the electric cur-
Driving Tents change the latter’s magnetism,
Pin the diaphragm or the armature is
2 moved accordingly. These move-
ments are imparted to the air.
If the horn is large enough the
resonance effect of the diaphragm
may be partially eliminated. Other-
Fig. 231—Horn type speaker wise when the frequency at which
mechanism. this diaphragm mechanically reso-
nates comes through the speaker
windings, a very loud output will be given out showing that
the loud speaker element is more efficient at this particular fre-
quency. The horn acts as a load upon the diaphragm much as a
resistance across the secondary of a transformer acts as a load on a
generator connected to the primary which supplies the power.
The power in this case is sound power radiated from the horn.
Very long horns which expand from a small opening near the
diaphragm to a very wide mouth in an “ exponential ’? manner
give the best response characteristic, in that they are freest from
resonances.
If the horn is replaced by a cone, as shown in Fig. 232, much
better frequency response results because the cone has a larger
Permanent Magnet
Non Magnetic
Diaphragm
THE MOVING COIL SPEAKER 385
area and can give appreciable sound output at low frequencies.
With a given diaphragm area, haly-
ing the frequency requires four times
the relative motion of the diaphragm as Paper Diaphragm
to produce the same sound power, i
while increasing the size of the cone
so that much greater areas of air are
displaced with a given amount of
diaphragm (cone) motion, enables
lower frequencies to be reproduced.
Such speakers have an impedance
that increases with frequency, being
about 1000 ohms at 100 cycles and
running as high as 40,000 ohms at
5000 cycles. This means that the
tube works into a constantly varying
‘impedance as the frequency varies, and that most efficient transfer
of energy, or transfer with least dis-
tortion, can occur at only a small range
of frequencies. Owing to the fact that
the impedance is low at low frequen-
cies, more distortion due to curvature
of the tube characteristic takes place
at this end of the audio band. It isa
fact that such speakers have a very
limited motion so that it is impos-
sible to transmit very large low-fre-
quency responses to them, and so the
distortion due to curvature is not so
pronounced as theory would indicate.
For average home use, a good cone
speaker about 2 feet in diameter will
satisfy the vast majority of listeners.
Fic. 233.—Modern “dynamic” 312. The moving coil speaker.—
or moving coil loud speaker. THe gonstruction of the moving coil
or dynamic type of speaker is shown in Fig. 233. A strong
magnet is energized by a direct current, either from a storage
Magnets
Connecting Rod
Fie. 232.—Cone type speaker.
Supporting Ring
for Base of Cone LN
>
TT 71 “Spool for
J Moving Coil
Leads to Output Coil A
of Power Stage Qutput
Transformer
386 RECEIVING SYSTEMS
battery, or from part of the plate current supply system, or from
the 110-volt a.-c. line by means of a rectifier. The voice frequency
currents coming from the final power tube in the amplifier are
passed through a few turns around a small movable coil to which
is attached the diaphragm, or cone. When a.-c. currents flow
through the coil, it tends to move at right angles to the lines of
force across the air gap. These motions of the coil are imparted to
the cone and thence to the air.
The impedance of this movable coil is very low, of the order of
5 to 10 ohms, and is almost constant at audio frequencies. This
means that the tube looks into its own impedance through a step-
down transformer. For this reason a much flatter response curve
is possible. Because the coil can move through a considerable
distance without danger of any mechanical noises—such as caused
by the diaphragm rattling against the poles of a unit of the type
of Fig. 231—much better low-frequency response is possible. Con-
siderable sound energy can be got from such a speaker. The reso-
nant frequency of the moving part is usually lower than the lowest
audio tone to be reproduced.
313. Baffles for dynamic speakers.—It is necessary to install
such a speaker in the center of a rather large and heavy “‘ baffle ”’
if the low notes are to be properly reproduced. Otherwise the wave
set up from the back of the cone can interfere with the wave set up
by the front with the result that little or no sound gets to the
listener. The baffle increases the air path between front and
back and should be great enough so that the _ shortest
mechanical path between front and back edges is at least
one-quarter wavelength for the shortest lowest note to be
received. Since the wavelength of sound, like that of radio
waves, is equal to the velocity it travels divided by the frequency,
it is not difficult to prove that a baffle at least 32 inches square
is necessary for notes as low as 100 cycles, and 110 inches for
notes as low as 30 cycles. When the unit is mounted in a box,
peculiar resonances are set up which spoil the good qualities of
the moving coil speaker.
The turns ratio of the input transformer is about 25 to 1 with
conventional speakers and tubes.
THE TELEPHONE RECEIVER 387
314. The electrostatic loud speaker—The sound output in a
moving coil speaker is due to current variations in a coil of wire.
In the electrostatic speaker, sounds are set up by changes in voltage
on two plates which are insulated from each other. In other
words an electrostatic speaker is a condenser consisting of two
insulated plates maintained at a high potential difference with
respect to each other. Across the plates are added the changing
potentials coming from the last amplifier tube. These changing
potentials cause a motion of one plate of the speaker with respect
to the other and thereby set up sound waves.
In practice a small receiving tube is used as a rectifier supplying
from 200 to 600 volts of partially rectified d.c. to the plate of the
speaker. The current requirements are very small, thus a small
tube may give long life. The variations in voltage are fed to the
speaker through an output transformer.
315. The telephone receiver.—Long before the day of the loud
speaker, the pair of head phones which could be strapped to an
operator’s ears was the translator of electric currents into sound
waves. Such a device consisted of a permanent magnet and a
coil of wire wound about one or both of the poles of the magnet.
A diaphragm was placed in the permanent field of the fixed magnet,
and also in the field of the coil through which the changing electric
currents passed. Why is the permanent magnet necessary?
Consider only the coil through which the a.-c. currents flow,
and the diaphragm. Any current flowing through the coil regard-
less of its direction would attract the iron diaphragm. If, then, a
sine wave at 500 cycles were put through the coil 1000 times a
second there would be a maximum of current through the coil,
and the diaphragm would be attracted toward it this many times,
each time being pulled back to its original position by a spring or
by its own resilience. In other words, a 500-cycle note would sound
to the listener like 1000 cycles; all frequencies put into the receiver
or head phones would be doubled in pitch.
This causes no great difficulty in code reception, but in voice
reception it would be strange indeed. If, however, the diaphragm
is always under some mechanical tension by being near a strong
permanent magnet, then smaller variations in magnetism, due to
388 RECEIVING SYSTEMS
changing a.-c. current, would cause the diaphragm to fluctuate
about this permanent pull as an a.-f. average. Then when the a.-c.
current was in such a direction that it aided the permanent mag-
net, the diaphragm would be pulled nearer the pole piece; on the
+20
Canical Horn
Transmission Units
Transmission Unit
ed Arinature
Fixed-Edge Cone
—20
100 500 1,000 5,000 10,000 100 500 1,000 5,000 10,000
Frequency, Cycles per Second Frequency, Cycles per Second
Fig. 234.—Limited response of Fic. 235.—Response characteristic
horn speaker. of good cone.
other half-cycles when the a.-c. current was in such a direction that
it opposed or weakened the permanent field, the diaphragm would
be released from its mean or average position. The diaphragm,
then, would reproduce the exact tone, and not double it.
316. Loud speaker measurements.—It is not difficult to measure
the performance of a loud speaker. The procedure is to hang a
thout Equalizer,
‘Transmission Units
e-Edged Cone
10 50 100 500 1,000 5,000 10,000
Frequency Cycles per Second
Fra. 236.—Wide range of tones covered by dynamic speaker.
calibrated microphone in front of the speaker which is actuated by
various tones of known amplitudes from an oscillator. The output
of the microphone is amplified and measured, and thus a curve of
output versus frequency may be obtained. Such curves are shown
in Figs. 234, 235, and 236.
LOUD SPEAKER MEASUREMENTS 389
Dynamic type speakers have a tendency to deliver more output
at high frequencies where the cone ceases acting as a plunger and
begins to act merely as a small cone. An electric filter consisting
of inductance, capacity, and resistance may be put across the
speaker to absorb this greater output and keep it from bothering
the listener. If, for example, the frequency is about 4000 cycles, as
in Fig. 236, a condenser in series with resistance and inductance
may be tuned to this frequency and placed across the loud speaker.
At all other frequencies it will be merely a high shunting path and
take no power. But at 4000 cycles its impedance is low, and cur-
rents of this frequency go through it instead of the speaker. The
resistance is used to keep the filter from being too sharply tuned.
Similar filters are used in phonograph reproduction to eliminate
the needle noise. They are called scratch filters and may tune
somewhere between 3000 and 5000 cycles. Of course such filters
take out their share of the desired signals and may make the
records sound “ boomy ” which is an indication of too much bass
for the amount of high notes.
CHAPTER XVI
RECTIFIERS AND POWER SUPPLY APPARATUS
Tuses which distort may act as rectifiers to transform a.-c.
currents into d.-c. currents and thereby be useful as sources of uni-
directional current either for charging batteries or for supplying
plate voltages to a receiver or other device. Such tubes have only
two elements, the source of electrons and the plate or receiver of
electrons.
317. The fundamental rectifier circuit—-When such a two,
element tube filament is heated to a proper temperature electrons
will flow to the plate provided it is at a
higher positive potential than the filament.
If an a.-c. voltage is connected between
the filament and the plate (Fig. 237) an
electron current will flow tothe plate when
the latter is positive and not when it is
negative. In other words current flows on
the halves of the cycle when the plate is
positive; on the halves of the cycle when
900000 the plate is negative, the electrons return
to the filament as fast as they escape. The
a.-c. voltage may be introduced into the plate-
10.287 Fundamental filament path by means of a transformer,
rechiner cacae which also supplies the heating current
for the filaments as shown in Fig. 238.
A d.-c. meter in the plate circuit would read a certain amount
of current which would be a value somewhere between the maxi-
mum current that flowed during each positive half-cycle, and zero.
The meter needle would not follow the rapid spurts of current and
so would assume some average value. If we consider as the input
390
THE FUNDAMENTAL RECTIFIER CIRCUIT 391
the a.-c. voltage, and the d.-c. meter as the output circuit, it is
clear that distortion is taking place, because there is no d.-c. in
the input and there is a readable amount of d.c. appearing in the
output. In other words the output is not a perfect replica of the
input.
Such a rectifier may be arranged to transform a.c. to pulsating
d.c. either at low, medium, or very high voltages. Whenever one
wants a source of d.c. voltage or current and an a.-c. voltage only
is available, a vacuum tube operating
as a rectifier may be employed.
Rectification takes place only in
this one direction. It is not a revers-
ible process. If we want a.c. from
d.c. we must use a motor-generator,
a converter, ora vacuum tube which
can be made to oscillate and thereby
convert a certain amount of d.-c.
power from batteries into a.-c. power. Fic. 238.—Operating rectifier
The amount of d.-c. current that filament from a. c.
would be read by a meter in the
plate circuit of such a tube would depend upon the voltage, the
form of the wave being rectified, the shape of the tube character-
istic, the amount of current that flows when the plate is negative
with respect to the filament, and upon other factors. If one could
listen to the output of such a tube, as one may by putting a loud
speaker in series with it, he would hear a buzzing or throbbing
sound.
The a.-c. voltage input wave is as in A, Fig. 239, which shows
the output without a rectifier. The pulses of d.-c. in the plate
circuit of a single wave rectifier are as in B. This is not a direct
current in the ordinary sense of the expression. It is a current
which varies in amplitude over the half-cycle of the a.-c. voltage
which makes the plate positive. The current throughout this half-
cycle flows in the same direction, however, and so may be con-
sidered as a pulsating direct current. These pulsations may be
smoothed out by filters and as a result nearly pure uni-directional
constant amplitude d.-c. current may be obtained.
Filament Winding
(Low Voltage)
\ Plate Winding
(High Voltage)
392 RECTIFIERS AND POWER SUPPLY APPARATUS
318. Kinds of rectifiers.—A rectifier, then, is a device which
transforms a.-c. current into a pulsating current which can be
smoothed out into pure d.-c. current if desired. Rectification may
take place (a) in a device which passes more current in one direc-
tion than it does in another, or (b) in a device which does not pass
any current at all in one direction, or (c) in a device which passes
Appearance it A. Voltage waves across
Output at X winding feeding tube
ap Load >X
No Rectifier
Etc, 2: Output of single wave
rectifier
Single Wave Rectifier
te
~ X 4 C. Output of double wave
Etc. rectifier
Double Wave Rectifie—
Fig. 239.
current when the voltage is increased beyond a certain limiting
value, but no current below that value or in the opposite direction.
If the plate of a rectifier tube is kept cool, so that it cannot act
as a source of electrons, the tube will pass no current when the plate
is negative with respect to the filament and the two-element tube
rectifier then is a member of class (b) above. If, however, a large
amount of rectified current is permitted to flow through a fairly
high resistance tube the plate may become hot, and so act as a
TYPICAL FILAMENT RECTIFIERS 393
source of electrons on the half-cycle when the plate is negative
with respect to the filament, and some “ back ”’ current will flow.
The tube then falls into class (a). Such a condition in a modern
plate voltage supply system is not likely
to occur and will be evidenced by a high
degree of hum in the output of the receiver.
Certain crystals, such as galena, silicon,
or silicon carbon (trade name “ carborun-
dum’), etc., are rectifiers, passing more
current in one direction than they do in
another and fall into class (a). Some
tubes do not use a filament, but have in
them a gas, neon or helium for example,
which ionizes when the voltage reaches a
certain value and conducts current in a
definite direction. They are members of class (c) and will be
described below.
The more perfect the rectification the greater will be the d.-c.
output from a given a.-c. input. If the rectifier is perfect and if
the input is a sine wave, a d.-c. meter in
a resistance load will read 0.901 times
the a.-c. input current. If 1 ampere a.c.
flows in, 0.901 ampere d.c. flows out.
319. Typical filament rectifiers.—
Tubes used primarily for rectifiers have
only two elements, the plate and the
filament. They are of two kinds, the
single or half-wave rectifier, and the
double or full-wave rectifier. The
Fic. 241—Current from a Single-wave rectifier has a single fila-
double- or full-wave rectifier to Ment and a single plate and rectifi-
the form approximated by J. cation takes place in it according to
the process described above. If desired
two such tubes may be arranged, as in Fig. 240, so that each
half of the a.-c. cycle is rectified, and so the output current
would look like Fig. 241.
A single-tube rectifier which operates on only half the a.-e.
Fie. 240.—Connections
of a full wave rectifier.
394
RECTIFIERS AND POWER SUPPLY APPARATUS
cycle is called a half-wave or single-wave rectifier. The two-tube
500
Plate Voltage
Plate Current per Anode
Plate Current Milliamperes
Plate Voltage
Fie. 242.—Characteristic of double
wave rectifier tube.
is called a full- or
It is
rectifier
double-wave rectifier.
| possible to combine both the
single-wave rectifier tubes into
one glass container by using
two filaments in series and two
plates. In such a case full-
wave rectification takes place
with the use of only one tube.
The 280 issucha tube. It has
two filaments and two plates
and is connected as in Fig.
239C.
Characteristic curves of
single- and full-wave rectifiers
are shown in Figs. 242 and
243.
320. Requirements for rectifier tubes.—The filament of a
rectifier tube must be rugged
and capable of supplying many
more electrons than are ever
needed for the proper operation
of the output circuit. Thus if a
tube is to supply 125 milliam-
peres steadily, it is possible that
in some circuits the instan-
taneous current through the
tube may be as high as 300
milliamperes and the tube
must be able to supply this
current without saturating.
If the tube saturates, the
pulse of current when the
plate is positive will be diffi-
cult to filter, and the recti-
Plate Current Milliamperes
Plate Voltage
Fig. 243.—Single wave rectifier tube
characteristic.
fier and circuit would suffer from other faults.
SINGLE-WAVE RECTIFIER 395
The resistance of the tube should be low so that no great
amount of voltage is lost in it, and so that the “ regulation ”’ of the
UX 281
Half Wave
|
| A.C. Input
eee
Rectifier Circuit
i}
|
|
Rectifier otk Filter
i
bine
Fig. 244.—Circuit of single wave rectifier, filter and load.
rectifier and filter may be good. A low-resistance tube of course
wastes less power, and so less heat must be dissipated. The insu-
lation between filament and plate must
be such that breakdown cannot occur
either due to direct puncture of some part
of the tube or to heating due to leakage
currents.
321. Single-wave rectifier.—Typical
single-wave rectifier circuits are shown
in Figs. 244 and 245. The only difference
between these circuits lies in the manner
in which the load is connected to the cir-
cuit, that is, whether it goes next to the
plate or next the filament of the rectifier
tube. In Fig. 245 the plate of the tube
is near ground potential but the trans-
former is ‘‘ hot.’’ In the other case, Fig.
244—-which is more generally used—the
Filament Plate
Winding Winding
Fig. 245.—Single wave rec-
tifier circuit employed by
some Western Electric
systems.
transformer is nearest the ground potential while the tube has a
396 RECTIFIERS AND POWER SUPPLY APPARATUS
high potential across it. Considering Fig. 244 when the plate end of
the secondary or high-voltage winding of the transformer is positive,
the other end of the secondary winding is negative with respect to
the plate end and a large voltage exists across it—perhaps 400 or
500 volts. This makes the end of the load attached to this end of
the transformer negative; and as we proceed in this direction,
through the load and back to the filament of the tube, the
circuit becomes more and more positive. The filament of a rectifier,
then, is the positive end of the circuit so far as the load is con-
cerned. The plate end is the negative or grounded end so far as
the load is concerned, and a voltmeter across the load must be
connected with this polarity in mind.
When the polarity reverses, on the other half-cycle, the plate
end of the transformer becomes negative and the filament end
positive. No current flows through the load or tube then, because
the plate cannot attract electrons to it from the filament. The
device is a half-wave rectifier only. It rectifies half the time.
When current flows, the voltage drop across the tube is only the
IR drop there—not the full transformer voltage.
Now suppose we connect another transformer secondary and
another tube as in Fig. 246. When the plate terminal of S,
attached to the plate of
tube 1 is positive; cur-
rent flows through that
+ tube and returns to the
center of the transformer.
Nothing happens in tube
2 because the plate of
this tube is negative.
When the direction of
the a.-c. voltage reverses,
Fic. 246.—Direction of flow of current in full the plate of tube 2 be-
Seyi TS. comes positive and cur-
rent flows through this
tube returning through the center of the secondary winding as
before. Current flows in a given direction through the load resist-
ance regardless of which tube is passing current.
GASEOUS RECTIFIERS 397
322. Gaseous rectifiers——An important rectifier is the Ray-
theon tube. It was the first really useful rectifier brought to
public use. It is a gaseous double-wave rectifier. The Raytheon
tube has undergone several improvements since 1925 when it was
first brought out and has been installed in many hundred thousand
plate voltage supply devices. It is one of the single most impor-
tant devices in the history of broadcasting.
The theory of the Raytheon tube is interesting. It employs no
filament, and therefore is not an electron discharge tube. To
understand its action we must go back
again to the electron theory.
Within the glass wall of the tube is an 4" 4
inert gas, that is, a gas which, like argon,
helium, neon, etc., does not combine
chemically with other elements. The mole-
cules and atoms of this gas are neutral
electrically, having neither a surplus nor
lack of electrons. There are, of course, a
few free electrons floating about within
the glass container, or a few may be re-
moved from one of the elements when a
voltage is placed on it. The elements
= “ A iti Negative
in a sinlge-wave rectifier are two: first, eee Electrode
a narrow strip or pillar of metal, and Fie’ 247 Raytheon tube
second, a larger envelope or plate of metal. construction’
If the tube is to be a full-wave recti-
fier, two strips are necessary. They are shown in Fig. 247.
Let us for the moment neglect the gas and consider a stray
electron that may be between the narrow and the large electrode.
Let us consider the moment when an a.-c. voltage on the two
electrodes makes the narrow electrode negative and the large plate
positive. The electron will be attracted toward the large plate
and will get greater and greater velocity the nearer it approaches
the plate. Since the large plate has considerable area it can take
care of all of the electrons that may be attracted to it. On the other
half-cycle the small electrode becomes positive, but because of its
limited area can take care of only a few electrons. The current
398 RECTIFIERS AND POWER SUPPLY APPARATUS
transported by these electrons is large in one direction, and small
in another, thus making the tube a rectifier falling into class (0)
of Section 318.
If the tube is well pumped and if the electrodes are properly
placed there will be little or no current flowing even though the
voltage applied to the tube may be considerable. Now let us put
some gas in the tube, helium for example. When an electron
finds itself in the field between the two electrodes it is attracted
toward the positive and gets up greater and greater speed as it
progresses. If the distance and the potential difference between
electrodes is great enough, the electron may get up enough speed
that when, and if, it hits a gas molecule it may knock another elec-
tron out of the molecule. This makes two electrons instead of one
and of course when they finally arrive at the positive plate they
deliver twice as much electricity as if only one were present. As
a matter of fact, when the second electron is knocked out of the
gas molecule it starts gathering speed, and before it reaches the
positive electrode may come in contact with another molecule and
release another electron. The process is cumulative, the more
electrons the greater the number of collisions per second, and the
greater the current carried across the space.
Now when the small electrode is positive, only a few electrons
can come into contact with it, and thereby give up their quota of
electricity, and the current during this half-cycle is small. The
swarm of electrons in the space between the two electrodes tends
to keep the current small because of the space charge effect.
There are two necessary features in this tube. There must be
gas which can be “ ionized ”’ by the electron collisions, and there
must be sufficient distance between the two electrodes for the elec-
tron speed to become great enough to dislodge additional elec-
tricity carriers. The voltage across the tube must be sufficient to
impart this speed to the electrons, and on one half-cycle there
must be some restriction to the number of electrons per second
that can arrive at the positive plate.
323. Characteristics of gaseous rectifiers.—The output of such
rectifiers is somewhat more difficult to filter than when an electron
flow tube is used because there is some back current, that is, current
THE TUNGAR RECTIFIER 399
which flows when the small electrode is positive. On the whole the
popularity of the gaseous rectifier tube has been not misplaced. A
typical Raytheon rectifier and filter assembly is shown in Fig. 248.
It differs from filament type circuits in the addition of two small
condensers across the two halves of the transformer secondary;
they are known as “ buffer ”’ condensers and usually are of 0.1 mfd.
Fig. 248.—Raytheon rectifier circuit.
capacity although more recent tubes require only about .01 mfd.
They eliminate any radio-frequency disturbances set up in the
circuit due to the abrupt manner in which the rectified current
starts and stops.
The Raytheon BH type has a rating of 125 milliamperes,
324. The Tungar rectifier.—The Tungar rectifier introduced in
1916 is a low-voltage, high-current tube of the combined gaseous
and filament type. It is designed to
rectify a.-c. current into a form suit-
able for charging batteries, and as a
matter of fact was used to supply cur-
rent for the filament circuits of some
early models of a.-c.-operated radio
Plate
Winding
receivers. Its starting or breakdown sci
voltage is about 15 volts and useful life A.C.
about 2000 hours. The gas is usually
argon.
A diagram of the connections of Filament
a typical tungar rectifier are shown in Pee gees ee
Fig. 249. It consists of a transformer nh ai
to reduce the input a.-c. voltage from '
115 to from 30 to 75 volts; the tube itself, which is a simple
400 RECTIFIERS AND POWER SUPPLY APPARATUS
two-element tube, consisting of plate and filament. The type in
general home use is a 2-ampere charger, which means that it will
put into a three-cell storage battery a current of 2 amperes.
Such chargers have been made to charge high-voltage batteries,
say up to 45 volts, but at a reduced rate of about one-half ampere.
The use of a.-c. tubes, however, will reduce the number of receivers
which use d.-c. tubes, batteries, and chargers.
Fig. 250.—A typical Tungar rectifier.
The theory underlying the Tungar rectifier does not differ
essentially from that already described for the filament and gaseous
types of tubes. The filament supplies the electrons which bom-
bard the inert gas and thereby produce more electrons and enable
such a heavy current as 2, or even 5 amperes in the case of larger
tubes, to be carried across the space between plate and filament.
A tube similar to the Tungar is the UX-866, which will supply
upwards of a half kilowatt without excessive voltage loss or
heating. Others will supply 30 kw. or more.
FILTER CIRCUITS FOR TUBE RECTIFIERS, FILAMENT TYPE 401
325. The copper oxide rectifier.—The copper oxide rectifier
1s outstanding among rectifiers by virtue of its simplicity and
reliability. The rectifier consists of a sheet of copper on one side
of which has been formed a coat of cuprous oxide (Cu20). Properly
made, this combination has relatively low resistance in the direction
oxide-to-copper, with very high resistance in the reverse direction.
The units are generally made in the form of washers, of 13 inch
outside diameter. These washers are then assembled in any
desired series and parallel arrangement on mounting bolts. Soft |
metal washers are placed between the oxide layer and the adjacent
metal surface for the purpose of improving the contact with the
oxide. The surface of the oxide is graphitized for the same reason.
This rectifier operates electronically and not electrolytically.
Rectification commences instantly on the application of voltage,
with no forming or transient condition interposed. Current is
carried not at points but uniformly over the available area.
Furthermore, the rectifying elements can be paralleled to any
extent. Operation in series presents no difficulties, as the discs
divide the voltage with approximate uniformity.
The outstanding feature of the rectifier is its long life. Units
on continuous duty life test, with battery load, show a reduction
of a little over 20 per cent in charging current in 33 years’ opera-
tion. This is remarkable performance for a rectifier, since it
operates entirely without attention or maintenance.
There are no limitations to the application of the copper oxide
rectifier. It may be used either half wave or full wave. The
bridge connection is commonly used since it simplifies the trans-
former design and furthermore permits units to be operated direct
from the a.-c. line without intervening transformer if so desired.
The quality of the d.-c. wave obtained is excellent so that filtering
is readily accomplished. Battery charging, battery elimination,
magnet operation, loudspeaker excitation, etc., in fact almost
any d.-c. application can be successfully handled by the copper
oxide rectifier.
326. Filter circuits for tube rectifiers of the filament type.—
There are several kinds of rectifier tubes as described in Section
319. The filament type is the simplest to understand and perhaps
402 RECTIFIERS AND POWER SUPPLY APPARATUS
the most commonly used and for this reason has been described in
detail.
The output of the rectifier circuit is not an even flow of current
at all. In a loud speaker it would make considerable noise, or if
used in a radio receiver without further alteration the hum would
be intolerable. The next step after rectification is filtering.
A good plate voltage supply device, then, consists first of a
transformer which raises the a.-c. voltage to the value required by
the receiver plus the losses in voltage in rectifier, filter and voltage
divider. Second, the rectifier which performs the task already
described. In the third place comes the filter whose task it is to
smooth out the pulsations of current in the plate circuit of the rec-
tifier so that the final product will be d.c. of constant amplitude
and a minimum amount of a.c. in it, and of a voltage high enough
to supply the voltage and current required by the receiver and
amplifier as well as the losses in the device itself.
A conventional filter circuit consists of series inductances which
smooth out the ripples of current and keep the current flowing at
the a.-c. voltage reversals, and shunt condensers which act as
reservoirs of voltage as described in Section 73. A two-section
filter is shown in Fig. 248, that is, two chokes and their accompany-
ing condensers. The amount of filtering necessary depends entirely
upon the amount of residual hum that is tolerable after the filtering
has taken place. A very quiet power supply device is required in
those receivers which have a rather high audio-frequency voltage
amplification and which amplify frequencies as low as 120 to
60 cycles.
There is always a certain amount of a.-c. voltage left after the
filtering has taken place. This voltage is a matter of millivolts
compared to several hundred volts of d.c., but even these small
a.-c. voltages may become objectionable when a loud speaker that
is efficient at low frequencies is used. This voltage is a combination
of the fundamental frequency, 60 cycles, and its harmonics. In
double-wave power supply devices the second harmonic or 120
cycles is particularly strong.
A good loud speaker which reproduces notes as low as 120 cycles
will hum badly when used with a power amplifier which gets its
REGULATION 403
voltages from a poorly filtered supply. Even with a very good
filter, some a.-c. voltages are likely to be picked up by the cores of
audio transformers if they are near power transformers carrying
a.-c. currents. The ultimate extent to which hum may be reduced
may be the amount that transformers can be shielded from stray
fields, and not the extent to which a rectifier’s output may be
filtered. The push-pull amplifier may be operated from incom-
pletely filtered supply. The rest of the current required from
the B eliminator can be carefully filtered at reduced cost.
327. Regulation.—After passing through the filter the circuit
returns to the transformer through the terminal resistance which
acts as a potentiometer to reduce the full voltage to the values
desired. If the full voltage is 180 volts, a tap at the proper
place will give 90 volts and another will give 45 volts for the
detector tube. Now it is apparent that the voltage across this
resistance depends upon the current through it, and if there were
no other resistances in the circuit Ohm’s law would tell us at once
what the voltage across the resistance would be. Unfortunately
the transformer, the rectifier tube, and the filter chokes all have
resistance, so that the greater the current taken from the whole
device the lower is the voltage across its output. From batteries
one gets 90 volts for his tubes whether he runs one or a dozen of
them; the drain from the batteries is all that changes. With a
voltage supply device, however, the voltage at the amplifier tap
would be less than 90 if the current taken from it exceeds a certain
amount. ‘This is a distinct disadvantage from every standpoint
but one.
In a high-resistance system it is difficult to blow up tubes. If
a B battery is put across a tube, the filament immediately becomes
very bright and probably burns up because too much current goes
through it. The current that can be taken from a high-resistance
device is limited and so it is doubtful if enough current could be
secured to damage the tube. This, however, is a rather dubious
advantage.
A series of curves showing the output voltage at various output
current drains gives what is known as the regulation of the device,
that is, the manner in which its voltage drops with increase in cur-
404 RECTIFIERS AND POWER SUPPLY APPARATUS
rent taken from it. The history of plate supply devices can be
traced in a record of the regulation of such units, the older they are
the worse their regulation curves, or the higher their internal
resistance. The steep regulation curve of an early “B eliminator”
may be seen in Fig. 251.
300
280
260}— yet
240 IL ee
220
200 + [
180
160 =
‘eee
a |1925"B Eliminator’
80
Output Volts D.C.
40 = LL (Se eel
40 4
20} +t
5 10 15 20 25 30 35
Output mA
Fig. 251.—Regulation curve of early ‘‘B eliminator.” Note the great
voltage drop.
328. A typical rectifier-filter system.—Several curves showing
the relation between the current and voltage in a modern rectifier-
filter are shown in Fig. 252.
The losses in voltage are those due the JF drop in the resistance
of the tube and the transformer. In Fig. 252 the voltage across the
output resistance of the filter would be less than these voltages by
the drop in the filter resistance. If the filter chokes have a d.-c.
resistance of 1000 ohms there would be an additional drop of 1
volt per milliampere of current from it.
A transformer which supplies 220 volts to each plate of a CX-
380 tube will deliver a voltage of 220 across the input to the filter
A TYPICAL RECTIFIER-FILTER SYSTEM
405
and if the latter has a resistance of 1000 ohms about 150 volts
will appear across the out-
put at a current drain of
65 ma.
The variations in trans-
former voltage, current
through the tube, and
steady load current are
shown in the oscillographs
in Fig. 253. The fact that
abnormally high instan-
taneous values of currents
must be passed by the tube
is clearly shown in the tube
current wave. No current
flows until the transformer
Effective D-C. Output Voltage
1100
Output of UX 281
1000 |_| Typical Recifier Circuit
of Rectifier
10 20 30 40 50 60 70 80 90100110120
Half Wave Rectification
Output Milliamperes
Fig. 252.—Output of modern rectifier.
voltage reaches a certain minimum value and current ceases to
flow as the voltage across the transformer secondary decreases.
The peak current rises as high as 310 milliamperes; and on the
Transformer’
Voltage
Fig. 253.—High instantaneous cur-
rent required from rectifier with
capacitive input.
assumption that the resistance of
the tube remains constant, the
power lost in it increases as the
square of the current, showing
that this high current puts a
severe burden on the tube. Asa
matter of fact the internal resist-
ance of the tube is almost con-
stant, decreasing somewhat at
higher current loads.
If the first filter condenser is
removed and placed across the
output, as shown in Fig. 254, not
only does the peak current passed
by the tube decrease, thereby de-
creasing the power lost in it, but the regulation is improved.
The disadvantage of such a connection is slightly greater
hum output and the fact that the output voltage is de-
406 RECTIFIERS AND POWER SUPPLY APPARATUS
creased by about 20 per cent. This loss in voltage may be
made up by increasing the secondary voltage of the transformer,
A.C.
Input
Plate Winding
Fig. 254.—Inductance input type of filter.
Filament Winding
and even then the power ‘osses in the tube will be less. The peak
current passed through the tube in the second case is about 140
milliamperes when a steady current of 125 ma. is required, thereby
reducing the power losses in the tube by about 35 per cent.
Figure 255 shows the oscillograph
Transformer studies of such a filter-rectifier sys-
Voltage VA tem.
The connection which omits the
first filter condenser increases the
life of the tube, permits the use of
a tube whose emission has fallen
below the point at which the tube
would be useless in a capacity in-
put filter, gives better regulation
at current drains. over 20 ma.,
reduces the filament emission re-
quired, and reduces the heating of
the tube.
329. Hum output.— Using chokes of 11.5 and 13.5 henrys re-
spectively for Li and Le and a 4-mfd. and an 8-mfd. condenser as
C;, and C2, a hum output of 44 millivolts is reported as a typical
case. Whether or not this amount of hum is objectionable depends
Tube Current
Fia. 255.—Reduced tube current
in an inductance input filter.
THE VOLTAGE DIVIDER 407
upon a multitude of factors including a matter of personal opinion.
The amount of hum that some listeners can tolerate would take
away all the enjoyment of radio from other listeners. Improve-
ment in loud speakers and amplifiers will make necessary greater
refinements in the way of eliminating hum, whether due to imper-
fect filtering, the use of a.-c. tubes or pickup from nearby wiring,
or iron cores carrying a.-c. currents.
Although the amount of hum that can be tolerated is a matter
of opinion, the following figures may be interesting. Using a sensi-
tive moving-coil loud speaker in a 3-foot baffle, a 120-cycle voltage
of 0.54 across the primary of the transformer feeding the loud
speaker was too loud for comfort. A voltage of 0.15 volt was a
desirable maximum. At 60 cycles these voltages were 5.2 and 1.3
respectively.
Hum measurements on twelve popular socket power devices
sold during 1927 gave the following r.m.s. voltages expressed as
percentages of the d.-c. output voltage. (To get the effective a.-c.
millivolts multiply by 10 times the d.-c. voltage.)
Load Current Minimum | Maximum Average
20 ma. from power tap...........- 0.007 0.67 0.025
AQ Mmastromy power tapes... sss: 0.009 1.0 0.05
ema e rOmrCebeCtOns ea mac eae a 0.001 0.15 0.015
Ey same). \orOvaN CIATKGHO > ooo ooanaecd oe 0.002 0.26 0.03
* 25 ma. load from power tap.
330. The voltage divider.—After the output of a rectifier has
been properly filtered it is necessary to feed it to the tubes which
need it. The power tube, or tubes, will require the greatest voltage
and the other tubes in the set will probably require 90 and 45 volts.
A voltage divider is necessary as a distributing system taking care
that each set of tubes gets its required voltage. It consists of merely
a resistor, or resistors, with taps at appropriate points so that the
full voltage is across the power tube and lower voltages at other
points in the system.
408 RECTIFIERS AND POWER SUPPLY APPARATUS
For example a rectifier-filter system may provide a total output
of 220 volts under load. This is the voltage measured between the
most positive and the most negative points in the circuit. Part of
this voltage, 40 volts for example, can be used to supply C' bias to
the power tube. This leaves 180 volts which may be applied
between the filament and the plate.
Let us consider the preliminary diagram in Fig. 256. Dis-
regarding for the moment the resistance across the B voltage
terminals, it may be seen that the plate of the tube gets its positive
potential by being connected to the positive end of the plate supply
unit. The negative
end of the unit is
connected to the
zs. center of the fila-
ment through a re-
sistance. The plate
current must flow
through this resist-
ance and there is a
voltage drop across
it. The grounded
end of the B plate
Fig. 256.—How the voltage divider originates. supply unit is the
most negative point
in the circuit. Going through the resistance toward the filament
and thence toward the plate through the tube we proceed steadily
toward a more positive point in the circuit. The filament end of
the C bias resistor marked (+) is at a higher or more positive
potential than is the grid end. The grid end, then, is negative and
the voltage drop across this resistor, due the plate current flowing
through it, may be used as the C bias for the tube. The voltage
actually between the center of the filament and the plate is the
total voltage across the plate supply unit minus the C bias voltage;
or in case a 171 is the tube, the plate supply unit must have a
terminal voltage of 220 in order to place a voltage of 40 between
grid and filament and a voltage of 180 between plate and filament.
It is now only necessary to place a resistance across the plate
B
Supply
C BIAS FOR TUBES OPERATED FROM A.C. 409
voltage supply unit so that various voltages appear along it, each
voltage being less than the voltage across the two ends. It is only
necessary to fix the values of the taps so that the voltages desired
are attained.
If another C’ bias is needed another resistor can be connected
to the negative end of the plate voltage supply and the filament of
the tube requiring the bias. The plate current of that tube flowing
through the new resistor to return to the filament, the source of
the electrons, will cause a voltage drop across the resistance the
negative end of which is toward the grid of the tube in question.
There is another method whereby the 40-volt bias for the power
tube is tapped and lower voltages are obtained for other tubes.
Such a method is liable to
lead to unwanted couplings
between tubes. For example,
in a radio-frequency ampli-
fier such a method of obtain-
ing bias may lead to regen-
eration or even oscillation
because the plate current of
some tubes flows through
the C' bias of other tubes.
Both these C bias con-
nections are shown in Fig.
257. One set is in series,
180
180
bas
NS
the other is a parallel set. Power Tube
The importance of by-pass- en
ing all such resistors has ¢ Bias
been mentioned in Section
233. A better method is Parallel’ BlasTans
to make the voltage di- yg, 257.—Both series and shunt C’ bias
vider part of the amplifier, resistances are illustrated.
as in Section 233. It is
only necessary then to supply one B plus voltage to the amplifier.
331. C bias for tubes operated from a.c.—When a tube is
operated from a.c. the voltage across the C’ bias resistor must be
slightly different than when the tube is operated from d.c. For
410 RECTIFIERS AND POWER SUPPLY APPARATUS
example let us consider the operation of a 171 tube from a storage
battery and from an a.-c. filament transformer. When the battery
|
40,5 >+<—5 >|
| \
!
iene ho Sa aaa
4
Fig. 258.—C bias
in battery operated
tube.
is used the 40.5-volt C battery is connected to the
negative filament terminal, which is considered
as the basis of reference to which all other volt-
ages are measured.
When the battery is replaced by an a.-c.
transformer, the basis of reference for both B
and C voltages is no longer one side of the fila-
ment because this is rapidly changing its po-
larity with the changes in polarity of the a.-c.
voltage. The new basis is the exact electrical
center of the filament, which is not far from the
mechanical center. The B and C voltages are
connected, not to one side of the filament but to the fictitious
center of the filament by means of a center
transformer or to a center tap on a resistor.
transformer is used with negligible resist-
ance in it.
In Fig. 258 is the d.-c. case in which the
filament resistance is represented as divided
in half and placed in each leg. The C
voltage is connected to the negative leg, and
the voltage measured from negative and plus
A are shown. Between the center of the
electron source and the point where the C
bias enters the system is half the filament re-
sistance across which is half the total voltage
drop or 2.5 volts, and so the C bias with re-
spect to the true center of the filament is 43
volts. When the C bias is introduced by
means of a resistor into the center tap of a
transformer (Fig. 259) the voltage drop across
this resistor—to take care of the 2.5-volt
tap on the lighting
Let us suppose a
Fic. 259.—C bias intro-
duced by plate current
flowing through resist-
ance.
drop across half of the filament—must be 43 volts. Then at the
two ends of the filament are voltages of 43 — 2.5 or 40.5 and
\43 + 2.5 or 45.5, giving as an average 43 volts as before.
ENGINEERING THE VOLTAGE
Now suppose we put a resistance across
center tap on it, as in Fig. 260. What
must be the voltage drop across the C
bias resistor to provide 43 volts to the
center of the filament? If this voltage is
43 volts, the voltage at the two ends of
the filament will be 43 volts plus and
minus the voltage drop in the resistance
across the filament. This voltage drop is
the plate current through the ohmic resist-
ance of the center tapped device. Thus if
the resistance is 100 ohms, and the plate
current is 20 milliamperes, the voltage
across each half of this filament will be 45
and 43, giving an average of 44 volts. Thus
it is seen that each new condition demands
some adjustment to get the required bias.
DIVIDER 411
the filament with the
Fig. 260.—Grid resistance
introduced into center of
filament by mid-tap re-
sistance.
332. Engineering the voltage divider.—The lower the resist-
ance of the voltage divider, the better will
be the regulation of
+
37.5+20
=57.5 |
mA
D
22.5+15
37.5mA
To Filter
and
Rectifier
Cc
20+2.5
=22.5
mia
R2
See)
Fig. 261.—Designing the voltage divider.
the entire device, but the
greater the load the tube
must bear. In general a
voltage divider is engi-
neered as follows. In Fig.
261, suppose the current
flowing through the resist-
ance R; is 20 milliam-
peres. This is known as
the ‘‘ bleed’ or “ waste”
current. It flows whether
or not there are any tubes
in the receiver that is sup-
plied with voltage from the
device. At the point B, a
voltage of 45 is desired. The
value of resistance is, accord-
ing to Ohm’s law, F + J and so is 45 + 0.02 or 2250 ohms. Since
412 RECTIFIERS AND POWER SUPPLY APPARATUS
this tap supplied only the detector plate circuit the current will be
about 2.5 ma., and is added to the 20 ma. taken by the lowest resist-
ance. Thus through Re flows 22.5 ma: and since 90 volts is desired
at point C the resistance Re will be 45 + 22.5 or 2000 ohms. If the
tubes which require 90 volts take a total of 15 ma. from the plate
voltage device, the final resistance will be 180 — 90 + 37.5 or
90 + 37.5 or 2400 ohms. The entire resistance will be 6650 ohms,
with taps at 2400, 2000, and 2250 ohms. The greatest amount of
power must be dissipated by the 2400-ohm resistor and so if the
entire resistance up to R4 is wound with wire large enough and on
a frame that can dissipate the heat corresponding to 4 watts,
there will be no trouble. Voltage dividers are available which
consist of a single winding of resistance wire on a heat-resisting
form. There are several sliders so that the correct voltages can
be obtained easily.
If any a.-c. current remains and flows through the voltage
divider an a.-c. voltage will be applied to the receiver to which this
part of the voltage divider connects. Hum results. If, however,
a condenser is connected from the high voltage end of this resistor
to the negative terminal of the divider, and if its reactance to the
a.-c. voltage is small the a.-c. currents will set up a small voltage
across the low impedance of the resistor shunted by a high capacity.
Problem 1-16. The maximum voltage needed with the voltage divider
of Fig. 261 is 180 volts, but the output of the filter is 200 volts under load.
What is the value of resistance R, to reduce this voltage to 180 and what must
be its power dissipation in watts?
Problem 2-16. Using a tube with a filament voltage of 7.5 volts and
operated from d.c. the proper C bias is 50 volts. The plate current then is
25 ma. What is the voltage that must appear across the C bias resistor when
the tube is operated from a.c. and when this resistor is connected to a center
of a resistanceless transformer winding? What must be the voltage if the
resistor connects to the center of a 100-ohm resistance?
Problem 3-16. What power in watts is dissipated in each of the resist-
ances in Fig. 261 under the conditions of Problem 1-16?
Problem 4-16. The resistance of the filter is 300 ohms. What must be
the output of the rectifier if the output of the filter is 250 volts and if 40 ma.
flow through the voltage divider resistance R4.
333. Voltage regulation.—In districts where the line voltages
vary considerably from hour to hour or from day to day, trouble is
VOLTAGE REGULATION 413
had with the poor regulation. When the line voltage is high, the
voltage is high, the voltages on the tubes will be high and their
life will be short. When the line voltage goes down the voltages
on the tube go down and the receivers do not work properly.
Several methods have been worked out to alleviate this difficulty.
One method involves the use of a line voltage regulator or
ballast lamp which passes 1.7 amperes (UX-876, C-376) at any
voltage between 40 and 60. The lamp is connected in series with
the primary of the power transformer. If the line voltage averages
115 volts, the load is so adjusted that 65 volts appear across the
primary of the transformer and 50 volts across the tube. Then
when the line voltage varies the voltage drop across the tube
varies and the voltage across the primary remains at 65 volts.
The chief difficulty with this tube is its slow action. It requires
several minutes for a steady state to be reached.
A tube that has been used much more than the ballast lamp is
the voltage regulator or “ glow tube,” the UX-874 or CX-374.
This tube is_ placed
across the 90-volt tap
in the voltage divider. 2 5 04-90
It is gas-filled and has a se —_
characteristic such that garrast is
the voltage across it is be
constant at 90 volts at
all currents from 10 to
50ma. The tube has
only two elements and
Glow Tube
the other two prongs of sae
the base are shorted Fic. 262.—Ballast tube to maintain constant
within the base. If one a.-c. voltage input to power transformer.
side of the a.-c. line is
connected to the socket into which this tube fits, the line will
be open unless this tube is in its socket and it thereby provides
a certain factor of safety. Both tubes are shown in the circuit
of Fig. 262.
The use of a simple series resistor between the line and one side
of the primary of the power transformer has been advocated and
414 RECTIFIERS AND POWER SUPPLY APPARATUS
practiced a great deal. It has only one virtue, the voltage on the
receiver can never get higher than a certain value which can be
made below the point at which tube life is threatened. If the
voltage goes down, however, nothing can bring back the voltage
across the receiver to its correct value unless the series resistance
is shorted.
The problem will probably resolve itself into one of making
tubes which can stand the voltage variations that are to be
encountered in practice, or in developing foolproof systems for
maintaining constant the voltage input to the receiver.
CHAPTER XVII
OSCILLATORS, TRANSMITTERS, ETC.
Because a tube acts as an amplifier, it can be made to generate
a.-c. currents of constant amplitude and at frequencies covering
the entire range from one or two cycles per second to well over
300 million cycles (one meter). Because the energy in the plate
circuit of the tube is greater than exists in the input or grid cir-
cuit, some of this energy can be fed back by several ways into the
input circuit and there amplified again. Starting with an initial
oscillation in the grid or plate circuit, provided the coupling be-
tween input and output circuits is of the proper phase and magni-
tude, this oscillation can be repeated and amplified, until finally
the tube maintains stable oscillations without the necessity of
exciting the grid from any outside circuit.
Before endeavoring to learn how a tube oscillates, etc., we
should get an idea of what the
term “ oscillation ’’ means.
334. Oscillating circuits.—
One of the most famous ex-
periments in all radio history
is that of charging a condenser
and letting it discharge through
a resistance and an induc-
tance. If the resistance takes
the form of a gap (Fig. 263) in Leyden Jar
the circuit across which a spark Fra. 263.—Familiar Leyden jar.
jumps, a photograph of this
spark made on a rotating mirror oscillograph shows that during
the instant of discharge the spark jumps back and forth across
the gap several times, first in one direction and then in another.
415
416 OSCILLATORS, TRANSMITTERS, ETC.
In other words the current in the condenser surges back and
forth across the gap instead of making one spark and thereby
thoroughly discharging the condenser.
Now such a circuit will produce an oscillatory spark in the
manner just explained if the resistance, inductance, and capacity
have correct values. When such an oscillatory spark takes place,
electric waves are set up in the ether. These waves have a fre-
quency determined by the well-known expression:
1
ae On LC.
~<— Amplitude ——>
—<— Amplitude —>
—-> Time Time
Fig. 264.—Highly damped Fig. 265.—Slightly damped (low resistance
series of waves. circuit) waves.
If the resistance is too great, the spark will not set up such waves.
The circuit is then non-oscillatory. If the resistance is decreased
to a very low value, the number of oscillations that take place
before the spark finally dies out becomes very great.
Thus in Fig. 264 the circuit has a high resistance; only a few
oscillations take place and these at a rapid reduction in amplitude,
each from the other. In Fig. 265 the resistance is very low and
many oscillations take place, the amplitude falling off slowly. A
circuit with much resistance is called highly damped because the
waves decrease in amplitude—are damped out by the resistance—
at a rapid rate. An undamped or continuous wave is generated
THE AMPLIFIER AS AN OSCILLATOR 417
by a theoretical circuit with no resistance, or one in which there is
a device which supplies the power wasted in the various
resistance.
335. Undamped or continuous oscillations.—With a given cir-
cuit, if the resistance can be reduced to zero through some means—
adding a negative resistance, for example, the damping factor due
the resistance is wiped out and the circuit generates continuous-
amplitude or undamped waves.
336. The amplifier as an oscillator.—Consider the box in Fig.
266, which is an amplifier. Any voltage put into it reappears in
the output magnified by the amplification factor of the device.
Suppose that the input is com-
posed of a coil and condenser,
and that part of the output =
voltage can be coupled to the
input coil. At the start sup-
pose this coupling coil T, com-
monly called a_ tickler, is
short-circuited, or removed
from the input coil. Now if py¢, 266—Essentials of an oscillator—
the condenser C is charged and an amplifier and feedback from output
then allowed to discharge sud- to input.
denly by closing the key, K,
oscillations will be set up which will die out at a rate depending
upon the resistance of the coil and condenser.
In the output circuit of the device will reappear an amplified
version of these oscillations. They too will die out. The energy
in them comes from some local battery, HZ, and the oscillations in
the input circuit only serve to release some of this local energy.
Now couple the tickler coil to the input in such a manner that the
voltage induced into the input coil by ordinary transformer action
is in phase with the oscillatory voltage. Then when the switch is
closed, a voltage appears across the input, is amplified in the
device, and in amplified form is impressed back on the input. This
will cause an increase in the oscillatory voltage, and so the effect
will be an ever-increasing series of oscillations, as in Fig. 267.
Ordinary oscillations are started in a tube circuit by thumping
418 OSCILLATORS, TRANSMITTERS, ETC.
the tube, or by turning on the plate battery, or by any sudden
change in the electrical or mechanical constants of the
circuit.
Oscillations in some circuits require appreciable time to build
up to their final value. For example a loud speaker which “ feeds
back ” mechanically into the elements of a detector tube may
finally result in a steady howl. If the loud speaker is near the
detector tube, or standing on the receiver cabinet, this chain of
events may occur. A sudden jar causes the elements in the detector
to change their relative position. This change takes place at an
audio rate depending upon
the natural mechanical fre-
quency of the element in
question. This audio tone
is amplified by the following
tubes and finally comes from
the loud speaker. The air
waves from the loud speaker
strike the tube and set the
—> Time elements into even greater
vibrations and finally the
Fig. 267.—Building up of oscillations in Whole system howls, or oscil-
resistanceless circuit or amplifier circuit. lates at an audio rate. Mi-
crophonic tubes, particularly
those with very small filaments, are prone to ‘ bongs”? which
may be amplified and lead to steady howls which take a second
or two to build up to final intensity.
337. Conditions for oscillation—Oscillations depend upon the
coupling between output and input, upon the fact that the device,
usually a tube, can amplify, and the fact that a combination of
inductance and capacity exists with resistance of such values that
the oscillations have the desired frequency. It is more difficult to
start and to maintain oscillations in a high-resistance circuit.
If the mutual inductance between grid coil and tickler is suffi-
cient to start oscillations, it can be loosened with an increase in
current. If the power in the oscillatory circuit is measured, it will
be found to go through a maximum when the effective resistance
4—Amplitude —>
MAXIMUM OSCILLATORY PLATE CURRENT 419
Ta
( a ) becomes equal to the tube plate resistance. If the C bias
on the tube is varied, it will be found that another value of mutual
inductance will be necessary to make the tube oscillate. All these
factors must be adjusted properly if maximum oscillatory power is
to be supplied by the tube.
338. Maximum oscillatory plate current.—The oscillating tube
may be thought of as an amplifier in which the exciting or input
voltage which is amplified comes from the tube itself; in other
words it is a self-excited amplifier. An alternating current flows
in its plate circuit just
as in any ordinary ampli-
fier. What is the maxi- 250
mum value of this cur-
rent? 200
Consider the tube :
working at the point A ya150
in Fig. 268. When oscil-
lations start, the a.-c. 100
plate current increases
from a small value dur-
ing the first few oscilla-
tions until it goes from
Zor £0 couple ae ee Fig. 268.—Manner in which oscillations begin-
value at A if a sine wave ning in tube circuit build up until entire
is being generated, or characteristic of tube is utilized.
until the plate current
curve flattens out. Then an increase in excitation does not
result in an increased plate current (a.c.). The limit has been
reached for the a.-c. plate current.
If the current at B is the saturation current, I,, of the tube,
the maximum a.-c. plate current will be
jie (1)
and because the a.-c. plate current is equal to the a.-c. grid voltage
420 OSCILLATORS, TRANSMITTERS, ETC.
multipled by the mutual conductance of the circuit, or 1, = Guy,
I,
E, a 2 G,,’
(2)
which is also equal to the voltage induced in the grid coil by an
a.-c. current flowing through the plate coil. This voltage is the
current in the oscillatory circuit multiplied by the mutual reac-
tance, or
E, = I1Mw (3)
from which we can calculate the current through this coil,
E, ‘e:
= —t = _ 4
Hs Mw 2GnMw (4)
“*__ (.m.s,) (4a)
ee, a
2V2GnM wo
Example 1-17. Suppose the saturation current of a power tube is 100
milliamperes (a low figure) and the other constants in Fig. 271 are L =
500 wh, r = 10 ohms, R, = 7150 ohms, G, = .76 X 10%, M = 160 ph,
f =100ke. What is the peak and r.m.s. oscillatory current, that is, the cur-
rent through the inductance, L, and what is the grid voltage due this current?
Solution.
I,
I =
2G
100 x 10s
~ 2X 6.28 X 100,000 X .76 x 10-8 x 160 x 10-6
= 0.655 ampere = .465 r.m.s.
Ey = I1p,Mw
c= Adsysy SIMO) SK WO SK OS OX IIOP
= 66 volts.
Problem 1-17. The voltage across the condenser in the tuned circuit of
the above example is equal to the current through it (which differs but little
from I7) times the reactance of the condenser at the resonant frequency.
Calculate the voltage across the condenser. The power used up in heating
the resistance of the coil is (Iz)? x r. Calculate this power (use the r.m.s.
value of Iz). Let this be called the useful power supplied by the tube. The
power from the plate battery is the product of the plate voltage and the plate
EFFECT OF COUPLING 421
current = I, X Ey. If the steady plate current is 50 milliamperes and the
efficiency of the system is 50 per cent, calculate the plate voltage necessary.
The efficiency is the ratio of the power supplied to the tuned circuit to the
total power supplied to the tube. Calculate this voltage.
Thus, E =, 1h 7 XG
Be = Ty,
Peet, scl,
P,
i = ==
Pr
339. Effect of coupling.—If such a circuit is set up in the
laboratory, it will be found that oscillations occur over a rather
wide range of coupling. Any one who has operated a regenerative
receiver for short, medium, or long waves knows that the loudest
signals are received just before the tube stops oscillating due to
too loose coupling between the secondary or grid coil and the
tickler.
Looking at formula (3) we see that the induced grid voltage,
due to the oscillatory current, is proportional to the coupling be-
tween the grid and plate coils. This induced voltage must be at
least equal to the original voltage there due to the condenser dis-
charge in order that the oscillations may be built up. If the
induced voltage is less than the original voltage, oscillations will
last longer than if the tube were not present, but they will finally
die out. The effect is as though we had reduced the resistance in
this oscillatory circuit but had not completely removed it. When
the induced voltage is equal to the original voltage, we have in
effect reduced to zero the resistance of the circuit, and oscillations
can keep up, although feebly. If, now, the induced voltage is
increased all of the losses in the input circuit will be made up by
the power taken from the local batteries (the plate battery) and
continuous oscillations will take place, gradually increasing in
amplitude until the entire characteristic of the tube is used.
If we start oscillations, and then decrease the coupling, and
if EZ, (induced) is to remain the same, the oscillatory current
must increase. Thus, when the coupling is loosened, but not
enough to stop oscillations completely, the oscillatory current
422 OSCILLATORS, TRANSMITTERS, ETC.
actually increases to make up the loss in induced voltage due to
the decreased coupling. When the whole
=
w
215
Siz ‘
=z i)
‘alo :
EIS ’
s J
I
plate current characteristic is used by
these oscillations, further decrease in
coupling must decrease the exciting grid
voltage, H,, and oscillations cease.
If, on the other hand, the mutual
Couplieg inductance is increased, oscillations rise
Fic. 269.—Effect of coup-
ling between in- and output
circuits.
to a maximum and then fall off and
finally cease entirely. The reason in
this case for cessation of oscillations is
different from the above reason. It is due to an increase in the
effective resistance of the tuned circuit, so that greater exciting
voltage is necessary to over-
come the increased losses.
The manner in which
coupling affects amplitude of
oscillatory current may be
seen in Fig. 269.
340. Dynamic character-
istics.—Because a change in
grid voltage produces a
change in plate voltage—just
as in a resistance-coupled
amplifier (Section 176), we
can not use the static charac-
teristic curves to predict the
action in the tube. We must
use the dynamic curves. In
Fig. 270 are the static char-
acteristic curves of a low-y
oscillator tube. When the
grid voltage increases in a
positive direction due to the
exciting or induced voltage,
100
SII oy
150 125 100 75 50 25 0
Eg (Negative)
Fig. 270.—Characteristic curves of power
oscillator tube. The dotted line is the
dynamic curve used when the tube oscil-
lates.
E,, the plate current increases but the voltage actually on the plate
decreases because of the greater JR drop in voltage across the
CONDITIONS FOR OSCILLATION 423
plate load. The operating point then may move along a curve
like the dotted line. Because of the decreased slope of the dynamic
curve compared to the static characteristic, the mutual conduc-
tance has decreased too, and it is not strictly correct to use the
static value in equations (1), etc. It is approximately correct,
however, unless the load in the plate circuit has a very high effec-
tive resistance, for example, a low-resistance circuit. In practice
this load is made up of not only the resistance always in the cir-
cuit but the resistance ‘“ reflected ” into it by transformer action
by coupling an antenna to the plate coil by means of a mutual
inductance and so has a fairly low effective resistance.
Modern tubes have such high values of saturation current that
they are never operated at the point at which the d.-c. plate current
is half the saturation value. Instead, they are so biased that the
average value is such that the power dissipated on the plate is
within the limits of safe heating. Then if sine waves are generated,
the maximum value of the a.-c. plate current is twice the value
read on the d.-c. meter. This value should be used in problems
and examples instead of the saturation value. For example a
UX-210 has a saturation current of an
ampere or more, but such a tube could
never be operated so that the d.-c. plate
current would be of the order of 500 ma.
Instead it is usually less than 100 ma.
341. Conditions for oscillation.—
When the tube and circuit oscillate we
can state that the resistance of the LCR
ircuit has been decreased to the point
that any oscillation starting there will
rot be damped out. In other words,
if the resistance of the circuit is R, We pig 271—Tuned plate cir-
must supply —R to it in order to get cuit (grid tickler) oscillator.
sustained oscillations.
In the circuit shown in Fig. 271 the value of resistance of the
circuit when coupled to the grid coil is
jpn sects!
Chew
424 OSCILLATORS, TRANSMITTERS, ETC.
Now all of the factors in this equation are positive with the
exception of M which may be either positive or negative depending
upon how L is coupled to the grid coil. When it is connected so
that oscillations occur, M is negative, and so the resistance A, in
the oscillatory circuit is decreased. If the coupling coil is reversed,
the total resistance in the oscillatory circuit is increased, and of
course sustained oscillations cannot be built up. By making
L, C, and M have the proper value, we can either add resist-
ance to the oscillatory circuit—and then no oscillations are pos-
sible, decrease the total resistance to zero, or make it negative.
L+ uM
The latter case is true when Se is greater than R.
Pp
The conditions for oscillation then are:
BBS ES Fy ek he
CR,
or
R L L CRR
M2=c— (x )- a eee
R i. Cit, m mM
or
mM L
—-~(M = Ch + —
i a Its
or
Gn M 2CR+ is
where the sign 2 means “ is equal to or greater than,”
and -+ means ‘ plus or minus.”
A number of facts can be gathered from these formulas. The
better the tube, that is, the greater its mutual conductance, Gm,
the looser can be the coupling and still maintain oscillations; with
a given tube whose mutual conductance is fixed, and with a given
coil-condenser combination, a certain mutual inductance is required
to start and maintain oscillations; the greater the resistance in
EFFICIENCY OF AN OSCILLATOR 425
the tuned circuit the better the tube must be with a given mutual
to maintain oscillations.
Problem 2-17. The mutual conductance of a power tube is 1500 micromhos,
its amplification factor is 7.6 and its plate resistance is 3500 ohms. It is desired
to generate oscillations of a frequency of 1000 ke. The coil to be used, in a
tuned plate circuit as in Fig. 271, has an inductance of 200 microhenrys and
a resistance of 10 ohms. Calculate the mutual inductance required to main-
tain oscillations. If the peak plate current is 100 ma.:—
What is the maximum current, Jz, that can exist in the oscillatory circuit?
If the power in this oscillatory circuit is I;? Xr and r is its resistance (10
ohms) what is the power dissipated there?
342. Efficiency of an oscillator—As in an amplifier, when the
grid is not excited, the power taken from the plate battery is equal
to I,E,, and this power is dissipated in heating the plate of the
tube. When oscillations take place the plate current and plate
voltage vary about their average or non-oscillating values. The
power taken from the battery does not change, but the power
wasted in heating the plate decreases, part of it going into the load
—yjust as in an amplifier (Section 189).
If the operating point is such that when the tube oscillates its
maximum a.-c. is twice the value read in a d.-c. meter and once in
each cycle is just reduced to zero plate current, the efficiency is
50 per cent. In Fig. 268 the average plate current is 0.125 ampere.
The minimum value it can reach is zero and the greatest value it
can reach is twice this value of 0.25 ampere. At the same time
the plate voltage variations are from zero to twice the average
value. In other words the variations in plate current are 0.125
ampere plus and minus 0.125 ampere, and the plate voltage is E,
volts plus and minus E, volts. This is an a.-c. voltage whose
maximum value is E, volts, and an a.-c. plate current whose maxi-
mum is 0.125 ampere.
The power, as in a resistive a.-c. circuit, is the product of the
effective current and the effective voltage, or
E, . I po E,Ly
Ne 2
and since the power supplied by the plate battery is L,I, one-half
the power taken from the battery is wasted in the tube and one-
426 OSCILLATORS, TRANSMITTERS, ETC.
half is used in overcoming the resistance losses in the oscillatory
circuit.
If we rate the efficiency of the plate circuit as the ratio between
the total amount of power taken from the B battery, [,H,, to the
‘ . ade Oe
power used in the oscillatory circuit, ioe we see that the above
condition represents an efficiency of 50 per cent.
If the C bias of the tube is so adjusted that the operating point
goes down on the lower bend, the steady plate current is small, and
the power taken
from the battery is
small. As shown in
Fig. 272, plate cur-
rent flows only when
the grid gets sufhi-
ciently positive to
permit it, that is,
when the operating
point gets up far
enough on the plate
current curve for
Fia. 272.—Form of plate current waves when tube current to flow.
is so biased that lower part of characteristic is used. Thus current flows
Grid Current
~~ Flows
during only a part
of the cycle, and because the a.-c. components decrease less
rapidly than the average value of current, the efficiency may
increase above 50 per cent.
343. Harmonics.-—If a large C bias is used, the form of the
oscillatory current is no longer a sine wave and of course many
harmonics are generated. Thus an oscillator generating a wave
form like that in Fig. 272 may be thought of as an oscillator pro-
ducing a pure sine wave plus many harmonics. If the oscillator
is used to supply power to an antenna which is tuned to the funda-
mental frequency, these harmonics put little power into this
antenna provided it is coupled loosely enough to the tuned circuit.
On short waves this is of great importance, because of the carry-
ing power of high frequencies. Thus if a tube oscillates at 40 meters
POWER OUTPUT OF AN OSCILLATOR TUBE 427
and puts out even a comparatively weak second harmonic on
20 meters, the latter signal can be heard over an area of many
hundreds or even thousands of miles. A broadcasting station
operating on 500 meters with a strong second harmonic can ruin
another station’s program operating on 250 meters, and so on.
There are times when it is desirable to generate a wave form
that has many harmonics, or a particular harmonic of large ampli-
tude. For example a quartz crystal is frequently used to control
the frequency of a transmitter. Because the thickness of the
quartz plate varies inversely as the frequency, for a high-frequency
circuit the crystal is very thin and there is danger of its breaking.
For this reason a thicker crystal is used and a harmonic of the
oscillator whieh it controls is used to drive a power amplifier
which works into the antenna. The crystal circuit, then, is an
oscillator which should generate a large second harmonic. By suit-
ably adjusting the C bias such an output can be attained.
Such a highly biased tube will have strong harmonics of much
higher order than the second, and if the frequency of the crystal is
accurately known these higher harmonic components of the output
of the tube can be used as standards of frequency over a very wide
range. It is frequently possible to count up to the 50th harmonic
of such a circuit. Thus if the fundamental frequency of the crystal
is 500 ke., the 10th harmonic would be 5000 ke., and the 50th
would be 25,000 ke.
344. Power output of an oscillator tube.—The power output
from an oscillating tube depends upon the efficiency of the circuit
and the amount of power that can safely be dissipated at the plate.
If the circuit is 50 per cent efficient and if the tube can safely
dissipate 50 watts on the plate, the output power is evidently
50 watts. If, however, a high C bias is used and a larger plate
voltage, the steady power taken from the battery may Increase
appreciably but a smaller proportion of it is lost in the tube and of
course more power into the load (the tuned circuit) obtained.
Thus if the tube is 70 per cent efficient, and can dissipate 50 watts
internally, the ouput power can be 116 watts and the votal power
supplied by the battery 167 watts. : :..
When an amateur manages to put into his transmitting tube
428 OSCILLATORS, TRANSMITTERS, ETC.
twice the power for which it is rated, he may still be operating the
tube as required by the manufacturer. He has increased the
efficiency of his circuit by operating at a high C bias, and by
making his output far from sine wave in form. The power lost
on the plate may still be within the manufacturer’s limit, and the
power obtained from the plate voltage supply unit and usefully
employed in putting signals into an antenna may be considerably
increased. If, however, his tube stops oscillating suddenly, due to
some maladjustment, the full plate battery power must be dissi-
pated at the plate and it is almost certain to be melted and the
tube destroyed.
Problem 3-17. An amateur desires to get 100 useful watts from a so-
called 50-watt tube. This means that 50 watts can be safely dissipated at
the plate. How efficient must his circuit be? If the plate voltage is 1000
volts, what will be the plate current? If his antenna has a resistance of
60 ohms at 40 meters, what antenna current must be put into it to radiate
100 watts?
Problem 4-17. The grid voltage necessary to excite a given trans-
mitting tube is 100 volts. The frequency is 500 ke., the tuned circuit induc-
tance, L», is 200 microhenrys, the coefficient of coupling between grid coil and
tuned circuit inductance is 0.3, the current in the tuned circuit is one ampere.
What must be the inductance L, of the grid coil?
Bie Malg =alfarx 10
Mowat (Dena 1 is
2 X 500,000.
ll
Ww
Problem 5-17. The resistance of the tuned circuit when coupled to an
antenna is 30 ohms. Its inductance is 300 wh, the uw of the tube is 8, its plate
resistance is 5000 ohms, the frequency is 300 ke. What must be the value of
M to make the circuit oscillate? M= oe R+ Z ‘
R CR,
Problem 6-17. The normal plate current of a UV 203—A tube is 125
milliamperes at a plate voltage of 1000. If the circuit is 65 per cent efficient,
how much can the plate current be increased at this voltage without using
more than 50 watts on the plate? What is the input and the output power
under these conditions?
It is possible to get circuits of high efficiency by increasing the
C bias so that plate current flows only during a part of the cycle
when the grid is positive. At these times the plate current is high,
OBTAINING GRID BIAS BY MEANS OF RESISTANCE LEAK 429
but at the same time the voltage actually on the piate is low—
because of the fact that the grid and plate voltages are 180° out of
phase and because of the high voltage lost in the load when the
current is high—and so the power wasted at the plate is low. If
the voltage at the plate could be reduced to zero no power would
be lost there and the efficiency would be 100 per cent. Such con-
ditions cannot happen, of course.
If the plate voltage is reducéd to a value comparable to the grid
voltage, the electrons from the filament would divide and more
would go to the grid than ordinarily, with the result that the plate
current would decrease. This would tend to increase the plate
voltage, and so the limiting condition of zero plate voltage cannot
be secured.
345. Maximum power output of oscillator.—As in an amplifier,
the maximum power is converted from the battery to the load when
the load resistance is equal to the tube plate resistance. Thus in
Fig. 271 the effective resistance of the load is L?w?/r which must
be equal to R, for maximum power output. This is not the
condition for maximum efficiency, but is the condition for maxi-
mum power output under a given set of conditions. As a matter of
fact the efficiency under these conditions is 50 per cent.
346. Obtaining grid bias by means of resistance leak.— During
the part of a cycle when the grid is positive (shaded area in Fig.
272) the grid draws current. When the grid is negative it takes
no current. There is in the grid circuit, then, an average grid cur-
rent. This current can be made to flow through a resistance and,
as in the case of a detector tube, be used to maintain the grid at a
negative potential with respect to the filament. Since grid current
flows it follows that some power must be wasted in the grid circuit.
This power is that wasted in the grid leak, usually of the order of
5000 to 10,000 ohms, and that wasted in the grid-filament resist-
ance. This power must be supplied by the plate battery, and
lowers the overall efficiency of the tube and circuit. If the current
is permitted to become very high the power dissipated in the grid
may become too great for the tube to handle, and breakdown
results. . :
In high-frequency circuits, the grid-filament capacity has low
430 OSCILLATORS, TRANSMITTERS, ETC.
enough reactance to conduct currents of considerable magnitude.
These currents must flow through the grid-filament input resist-
ance, and this represents another loss in power which must be kept
below the value that is safe for the tube. A choke coil placed
near the grid leak is one way to prevent unwanted oscillations
occurring at a very high frequency
partially determined by the tube
capacities. (See Fig. 273.)
Problem 7-17. The grid bias required
on a tube is 60 volts. If the bias resistance
is 5000 ohms, what must be the grid current
flowing? What power must the resistor be
capable of dissipating as heat?
R.F.C.
347. Practical circuits.—There are
a number of circuits which with an
Fie. 273. — Hartley oscillator; amplifying tube will produce oscilla-
grid current through &, provides tions, that is, transform d.-c. power
negative bias for the grid; choke +47, » battery or plate supply system
R.F.C. prevents high-frequency . :
Samiletions: into a.-c. power. All that is neces-
sary is a tube that will amplify,
coupling between input and output of proper phase and magni-
tude, and of course the filament, grid and plate power.
The coupling between input and output can be through induc-
tance, mutual inductance, capacity, or resistance, or through the
plate-grid capacity of the tube itself.
348. Hartley oscillator.—The simplest circuit, that is, the cir-
cuit which requires the least amount of apparatus, is the Hartley.
It requires only a coil with taps on it, a tuning condenser, the
tube and power supply. The coupling between plate and grid
circuit is through mutual inductance between the two parts of
the tuning coil, L, and L,. If the tuning condenser is placed
across the plate circuit only, the circuit is exactly the same as
Fig. 271. The circuit is given in Fig. 273. A.-c. currents flowing
in the plate coil, L,, induce voltages in L, which are applied to the
grid, amplified, and again applied to the plate coil. These voltages
are 180° out of phase because they are at opposite ends of the coil
with the center grounded to the filament. In Fig. 273 the plate
SHUNT-FEEDING OSCILLATORS 431
battery is placed in the center-tap so that it is at ground potential.
In Fig. 271 the plate battery, which has a high capacity with
respect to ground, is connected to the plate and thereby partially
shorts the plate of the tube so far as radio frequencies are con-
cerned. A better way is to use the circuit in Fig. 273. Since this
circuit would make the grid and plate at the same positive poten-
tial, which is the potential of the B battery, the grid is isolated so
far as d.c. is concerned by the blocking condenser. The proper
bias voltage is secured through a grid leak to the filament or
through a choke and C battery as in Fig. 274. The feed-back
between grid and plate circuits is adjusted by varying the center
filament tap. If more turns are in the plate coil a greater voltage
will be induced into the grid coil (L,I) and so greater feed-back
from plate to grid circuits would result.
For covering a wide range of frequencies with plug-in coils, the
Hartley oscillator is useful. It is used most frequently in laboratory
apparatus.
349. Shunt-feeding oscillators——In Fig. 273 the B battery is
in series with the plate coil. The terminals of the condenser, so
far as d.c. is concerned, are at the same voltage as the B battery
and of course this is exceedingly dangerous since the operator, who
is standing on the ground to which es ai
minus B is attached, may touch the
plates or shaft of the condenser, and
thereby provide a short within him-
self for the full plate voltage.
To prevent trouble of this kind
the plate voltage may be fed into the
tube through a separate path by
means of a choke coil and blocking
condenser as shown in Fig. 274. Now
the d.-c. potential of the plate is kept Fic. 274.—Shunt-feed Hartley os-
from the tuning coil and condenser cillator keeps high d.-c. voltage
by the condenser C. The reactance rom tuning condenser plates,
of the condenser must be low com- 5
pared with the reactance of the choke at the desired frequency SO
that the plate coil is not short-circuited so far as r.f. is concerned.
432 OSCILLATORS, TRANSMITTERS, ETC.
In a similar manner the use of a blocking condenser in the grid
circuit and a choke for feeding the C bias into the grid is a shunt
or parallel feed method of separating the d.-c. and a.-c. currents
and voltages. In Fig. 274 there are no d.-c.
currents or voltages on the tuning circuits,
nor a.-c. currents in the C or B batteries.
350. Other oscillating circuits.—If the
inductances and condensers in Fig. 273
are interchanged, as in Fig. 275, we have
the Colpitts circuits, a typical arrange-
ment of which is shown in Fig. 276. In
amateur practice the tuning condensers
are on the same shaft which is grounded to
the filament. The chief advantage of this circuit lies in the small
a.c. potentials across the chokes.
In Fig. 277 is the tuned-plate-tuned-grid circuit. Input-output
coupling is provided by the tube’s grid-plate capacity—one of the
few places in radio circuits where this
unwanted and obnoxious capacity is put
to use. Whenever the plate circuit is
tuned so that it is sufficiently positive in
reactance, the system will oscillate, and
it is not when the plate and grid are
both tuned to the same frequency as
many amateurs think. In this case the
plate inductance must always be some-
what greater than that necessary to
resonate the tuning condenser to the fre-
quency at which the grid circuit is
tuned. Tuning the plate circuit to a Fre. 276.—Practical Col-
high wavelength, or lower frequency pitts circuit.
than the grid, is the same thing said in
other words. This circuit has somewhat greater frequency
stability because of the high-impedance plate load. In practice
the C bias would not be shunt fed.
A resistance feed-back that is employed in laboratory oscilla-
tions is shown in Fig. 278. As in all such circuits, the grid excita-
Fig. 275.—Fundamental
Colpitts circuit.
ADJUSTING THE OSCILLATOR 433
tion is greatest when the largest coupling between grid and plate
coils is used. Then the plate circuit variations are considerable in
magnitude, they use curved parts of the characteristic, and har-
monics are generated. In a laboratory oscillator where harmonic
production is to be kept to a minimum, the coupling between input
and output circuits should be adjusted at each frequency to the
least possible amount that will insure stable oscillations. The re-
sistance feed-back method is useful in such oscillators because of
the mechanical ease of adjusting the feed-back voltage.
In Fig. 278 the coils are iron core inductances and the currents
generated are at audio frequencies.
Cg P
=e
1
i
ALF.
Lp Choke
Fig. 277.—In the tuned-plate tuned- Fia. 278.—Cireuit often used
grid circuit Cy, acts as the feed- in low (audio) frequency os-
back agency. cillators.
351. Adjusting the oscillator— When the oscillator is used to
deliver power to an antenna it is desirable to attain the adjustment
which will either deliver maximum output, or secure maximum
efficiency so that greater inputs may be used. The tuned-plate
tuned-grid circuit has no adjustments. The operator tunes either
the plate or grid circuits until the tube oscillates and there is noth-
ing else he can do. In fact, if he does not tune the circuit to an
oscillating condition the plate current may be very high.
In the Hartley circuit, however, it is possible to move the center
filament tap and so to get some control over the strength of oscilla-
tions, the feed-back voltage, etc. In Fig. 279 is illustrated the
result of varying the filament tap on a simple 40-meter oscillator
of the Hartley type. The curve gives the plate current, I,, the
1434 OSCILLATORS, TRANSMITTERS, ETC.
current, Ja, into an antenna coupled to the plate coil, and the ratio
of antenna current to plate current as some measure of the effhi-
2 3 4 5
Turns in Grid Circuit
Fic. 279.—Effect of varying grid turns,
thereby changing excitation.
ciency of the circuit.
What is wanted, from
an amateur’s standpoint,
is much antenna current
and little plate current.
The greater is this ratio,
the greater is the effi-
ciency of his circuit.
The connection be-
tween plate voltage and
oscillating current is
shown by Fig. 280 to
be linear. The grid bias
measured across a 5000-
ohm resistor is plotted
too. The tube was a UX-—210 oscillating at 1225 meters in a
tuned plate circuit, Fig. 271. The effect of changing the C bias
resistor of a tube
in the tuned-grid-
tuned-plate circuit
is shown by Fig.
281.
’ 352. Frequency
stability. — When
such circuits are
used for transmis-
sion or for labora-
tory measurements
where a constant
50
75
TOO 125 TSO 175200225
E, Volts
frequency output Fic. 280.—Relation between plate voltage and
is desired, compli-
cations set in. The
oscillatory current and grid bias.
frequency of such circuits is determined chiefly by the induc-
tance and the condenser across it.
shunted by the input capacity of the tube. This input capacity
This condenser is also
FREQUENCY STABILITY 435
is a function of the plate load and the grid-plate capacity of
the tube. In fact the input capacity C; as a function of the
load and this grid-plate capacity is
R
Cra itil ee )
aes ie oP
C002
000%
3 3
So [=]
o Oo
000FT
O009T
0008T
00002
00022
000r2
00092
=} S
Grid Leak Resistance Ohms
Fic. 281.—Effect of varying grid leak resistance.
which shows that any change in the plate resistance R, of the tube
or in the grid-plate capacity of the tube, or the output load Ro
produces a change in the grid-filament capacity which may have a
share in determining the frequency to which the system oscillates.
Changes in filament temperature, in C bias, or in plate voltage will
affect the plate resistance of the tube and change its relation to the
load resistance. Such changes produce a change in frequency of
436 OSCILLATORS, TRANSMITTERS, ETC.
the oscillator’s output. The smaller is the plate load, the larger
the grid-plate capacity; the greater the plate resistance of the tube,
the more will the generated frequency depend upon these factors.
One way to lessen this difficulty is to make the fixed input
capacity of the tube which is across the tuned circuit so large com-
pared to its normal grid-filament capacity that changes in the latter
are unimportant. Thus, shunting a fairly large capacity directly
across the grid and filament will increase the total effective input
capacity so that small changes in the internal capacity of the
tube will have little effect upon the tuning.
Another method is to make the tuning condenser very large, in
other words to use a high-capacity-low-inductance circuit. Such
circuits have large circulating currents in them, but small voltages,
and at times are very inefficient.
353. Master oscillator systems.—Where large amounts of
power are to be transferred to an antenna, or other load, the place
of the single oscillating tube is taken by a smaller oscillator and a
large power amplifier driven by this tube. In other words we have a
self-excited oscillator and a separately excited amplifier. Ifthe oscil-
lator is carefully stabilized against frequency changes, the output
of the amplifier will be constant too. Changes in the load (the
antenna for example) into which the amplifier works will not affect
the frequency at which the oscillator is generating.
Such a system is called a ‘‘ master oscillator, power amplifier ”’
system and is used in all broadcasting stations and all transmitters
of appreciable power. The oscillator can be of any conventional
type; the amplifier may be a single tube, may be several in parallel,
or push-pull, or several in cascade, just as in audio amplifiers. The
chief difference lies in the fact that considerable power is being
handled and so the circuits are made up of heavy conductors and
use large water-cooled tubes which may use plate voltages as high
as 10,000 volts and several amperes of plate current. (See Fig.
282.)
The amplifiers may be neutralized—usually are—or they may
use screen-grid power-amplifying tubes which have very low grid-
plate capacities and so need no neutralization.
A simple master oscillator system is shown in Fig. 283. Here
CRYSTAL CONTROL APPARATUS 437
the power amplifier is a 50-watt tube which is to be operated at
50 per cent efficiency. That is, 50
watts go into the load, and 50 into
the tube. If the plate voltage is
1000 and the instantaneous voltage
on the plate is to be just reduced to
zero once in each cycle, the peak a.-c.
voltage will be 1000. If the u of the
tube is 25, the a.-c. grid voltage
must be at least 40, and if the in-
ductance of the oscillator plate cir-
cuit is known it is a simple matter
to calculate the current that must
flow through it to set up a voltage of
40 with which to drive the following
tube.
354. Crystal control apparatus.
—A quartz crystal cut with proper
respect to its optical axis has the
ability to control the frequency
al Inlet
Fig. 282.—Construction of water-
cooled tube.
of an oscillator to a remarkable degree. When such a slab
of quartz is compressed mechanically, an electrical difference of
Oscillator
oe
1
|
|
|
|
Amplifier
R.F. Choke
+B
Fic. 283.—Master oscillator-power amplifier transmitting circuit.
potential is generated across its two faces. Conversely, when such
a difference of potential is set up across its faces, it tends to change
438 OSCILLATORS, TRANSMITTERS, ETC
its size. It may be thought of as a tuned circuit whose frequency
is fixed by the dimensions of the crystal. The thicker it is the
longer the wavelength to which it resonates. The manner in which
the crystal is cut is of utmost im-
portance; if it is cut in one man-
ner, the relation between frequency
and thickness is 2.64 meters per
thousandth inch, and if cut in
¢ another, the relation is 3.87 meters
Fra. 284.—Place of crystal in os- per thousandth inch. A crystal
eillatad crouse: quartz plate 1.0 millimeter thick
will resonate to 2860 ke.
If connected as in Fig. 284 it takes the place of the tuned grid
coil in the tuned-plate tuned-grid circuit. For several degrees of
the plate tuning condenser the output frequency is that of the
crystal, and so changes in plate resistance of the tube, battery
voltages, ete., will have a relatively small effect on controlling the
frequency at which the cir-
cuit oscillates.
The frequency at which
the crystal resonates depends
to some extent upon the
temperature of the crystal.
In the best transmitting
stations, the crystal plate is
maintained at constant tem-
perature by complicated
thermostats and_ electrical Temperature Degrees C
heating coils. Then the fre- Fic. 285.—Effect of temperature on fre-
quency of a_ broadcasting quency.
station can be maintained
within 50 cycles of its assigned frequency, say 1000 ke. This is an
accuracy of 50 parts in one million of 0.005 per cent. The effect
of temperature on a small oscillator may be seen in Fig. 285, and the
effect of changes in plate voltage on the oscillator tube in Fig. 286.
355. Frequency doublers.—The amount of power a crystal can
control is limited. If the crystal is called upon to handle too much
AZZ
Crystal
Change in Frequency
Cycles per Second
FREQUENCY DOUBLERS 439
power it is liable to crack. The power that can be controlled is
about the output of a 5-watt tube on broadcast and higher fre-
quencies where the plate becomes very thin, and certainly not
over 50 watts can be controlled with safety.
The crystal oscillator is followed by amplifier stages until the
required amount of power is ready for the antenna. These stages
of amplification can be single tubes neutralized if necessary; they
may be push-pull tubes, they may be tubes in parallel. The pur-
pose of each succeeding tube is to provide a voltage at the required
frequency and of the required
magnitude which may be used pea
to drive the grid of the next
tube.
For high frequencies the
problem of crystal breakage,
and from oscillation in the 60
amplifier stages, becomes seri-
aoe ye ets ee Fig. 286.—Effect on frequency of changes
followed is to use a fairly in plate voltage of a crystal-controlled
large crystal which drives an P ecilator
oscillator. Let us suppose
the antenna is tuned to 40 meters. The crystal may oscil-
late at 160 meters. The tube is so biased that it generates
a strong second harmonic, or 80 meters. This 80-meter out-
put is fed into the grid of the next tube, whose output may
be tuned to the second harmonic of 80 meters, or 40 meters.
Neither of these amplifiers shows much tendency to oscillate
because of the fact that the input and output circuits are tuned
to different frequencies. By proper values of C’ bias a strong
harmonic can be secured and the final power tube driven by a
40-meter voltage.
Such transmitters can operate from either a.c. or d.c. If they
are to be used for telephone transmission, broadcasting, for exam-
ple, pure d.c. is necessary. For code transmission some slight
ripple in the output is probably desirable if the signals are to be
copied by ear at the receiving station. If they are to operate a
relay and mechanical apparatus, the character of the signal may
Nh
o
—~
(=)
Change in f
Cycles per Second
eae
50 100 150
EpVolts
440 OSCILLATORS, TRANSMITTERS, ETC.
be adapted to the receiving apparatus, or ignored entirely. A
‘frequency doubler ”’ is shown in Fig. 287.
356. Self-rectified transmitters—When a.-c. voltages are
to 2f
Antenna Tuned
A second harmonic (2f) voltage in the plate circuit of the
Amplifier Neutralized
and Tuned to 2f
Fic. 287.—A frequency doubler.
oscillator drives the following amplifier.
applied to an oscillator tube,
it rectifies them and plate
current flows during the time
the plate is positive. In other
words, the circuit oscillates
half the time. On the nega-
tive half-cycles the circuit is
non-operative. The signal as
it is heard at the receiving
station has a _ characteristic
note depending upon the fre-
quency of the transmitter
power supply, and the ad-
justments at the transmitter.
A transmitter of this type is
called a self-rectified circuit
because the tube furnishes its
own plate voltage by rectify-
ing an a.-c. wave. Two such
tubes may be used in ~push-
pull or “back to back” to
rectify and oscillate on oppo-
site halves of the a.-e. cycle.
The transmitted note then
will have double the frequency
of a single-wave _ rectified
transmitter. A circuit of this
kind is shown in Fig. 288.
Such transmitters take up
more room in the ether than is
desirable, and even though controlled by crystal seem to vary
from their assigned channel because of variations in the audio-
frequency modulations. A transmitter using 500-cycle source
of plate voltage will require a channel width of 1000 cycles
ADJUSTING THE PLATE LOAD TO THE TUBE 441
when holding its key down. Ether space required for it depends
upon adjustments of several circuit factors.
357. Adjusting the plate load to the tube.—The condition for
maximum output of a tube with a given amount of power to be
taken from the plate battery is that the load into which the tube
works is equal to the plate resistance of the tube. The load is the
resistance of the tuned circuit. Now the value of L and C are more
or less fixed in this circuit, which leaves the series resistance as the
only independent variable factor. Suppose, for example, we desire
to generate oscillations 1000 ke. in frequency in a tuned circuit.
This determines the frequency. We can choose L and then C is
110V
60~
Fria. 288.—Full-wave self-rectified oscillator operated entirely from a.c.
fixed; we cannot change their ratio which determines the effective
resistance, L/Cr. What can be done so that the tube generates
the maximum amount of power in the tuned circuit?
If the proportion of the entire tuned circuit across which the
tube is connected is varied the impedance into which the tube works
will be stepped down; and by properly choosing the plate tap in
Fig. 289 the proper load will be presented for the tube to work into
so that the maximum power will be delivered. The inductance of
the tuned circuit may be looked upon as a transformer which
couples the tube to its load, usually an antenna.
358. Plate current when oscillator is connected to antenna.—
An antenna-counterpoise system is usually connected to the power
tube through an inductance which is a part of the tuned antenna
442 OSCILLATORS, TRANSMITTERS, ETC.
system. This antenna system is tuned to resonance with the tuned
circuit. If the oscillator is adjusted to its proper frequency by
varying the inductance or capacity of its tuned circuit, and the
plate tap is then adjusted for maximum power output, the antenna
coupling inductance may be brought near the tuned circuit induc-
tance. If then, the tuning capacity of the antenna is varied
through resonance with the transmitter, current will begin to flow
into the antenna, and the plate current of the tube will probably
increase because the battery must now furnish the power taken by
the antenna as well as the power lost in the tuned circuit. Closer
coupling will induce a greater voltage in the antenna, more antenna
current will flow, the plate current will increase, and greater power
Key in
Grid Leak
tape
500
2.0
Mfd (a)
Fie. 289.—Maximum power is ob- Fig. 290.—Two methods of keying
tained by adjusting tap until tube transmitter. Note “thump” filter
works into its own impedance. in —B lead.
will be radiated. Since the antenna is being coupled to the tuned
circuit its resistance is being reflected into the plate circuit and
some readjustment of the plate tap must be made. This may
change the frequency slightly, and so all four variable factors—
coupling, plate tap, and tuning, and tuned circuit capacity—must
be adjusted until the maximum power is transferred to the antenna
at the required frequency.
A rough idea of the power being radiated may be estimated
in the foliowing manner. Suppose the plate current without the
antenna is 100 milliamperes and the plate voltage is 1000 volts.
This represents a power input to the tube of 100 watts. Now
suppose the antenna is coupled to it, and the plate current increases
to 150 ma. The power is now 150 watts. The difference between
150 and 100 watts, 50 watts, may be assumed as going into the
KEYING A TRANSMITTER 443
antenna. If the antenna current is measured, a rough estimate
of the antenna resistance may be had. The method is not at all
accurate unless a sine wave is being radiated and then only
approximately.
359. Keying a transmitter.—There are several methods of
modulating the oscillations of a transmitter so that they may con-
vey intelligence from one operator to another. The tube may be
caused to cease oscillations, the antenna circuit may be broken or
closed or the frequency to which the oscillator is tuned may be
changed in accordance with the manipulations of the key. A key
placed in the plate voltage supply (Fig. 291) will cut off the power.
This method is not successful with well-filtered systems because of
uv
Cc
Oo —
32 100V
a 60~
o
a
; ; Key in Power
Key in Mid Tap Transformer Primary
Fic. 291.—Keying in ‘‘center tap”’ or in transformer primary.
the time taken to completely discharge the condensers. An alterna-
tive place for the key is the grid circuit, as in Fig. 288 or 290. If
the grid leak or the C bias lead is opened by the key, there will be
no path for the electrons trapped on the grid to escape. The grid
will them assume a large negative voltage which will reduce the
plate current to such low value that oscillations cease.
If the key is placed in the plate lead it must break the full
power to the tube, and its contacts must be able to handle the cur-
rent without heating and without breakdown from the voltage
drop across the key as it is opening.
The voltages and currents in the grid circuit are much lower.
Both of these methods of starting and stopping the oscillations
are abrupt and provide the nearby ether with profound shocks
known among the amateur fraternity 4s “key thumps” and
444 OSCILLATORS, TRANSMITTERS, ETC.
cordially hated by listeners to other transmissions. Various thump
filters have been devised to start and stop the oscillations in the
tube less abruptly. One method is to let the tube oscillate feebly
during the periods the key is up by placing a high resistance across
the key contacts. Another is to place a resistance, capacity, and
inductance across the key contacts. The condenser charges and
discharges slowly and prevents the abrupt opening of the plate
power circuit. The inductance (about 0.5 to 2 henrys) slows up
the start of oscillatons.
Keying can be accomplished as in Tig. 291 but generally the
break is made between the filament and the connection to minus
B, which is also attached to the grid. In other words the filament
only is cut loose. This is called ‘“ keying in the common lead ”
and is quite common.
360. Too close coupling to antenna.—If the closed circuit is
properly tuned and the antenna coupled to it, it may be found that
the emitted wave is no longer of the desired frequency or that the
tube refuses to oscillate or that as the antenna tuning condenser
is varied two maxima of antenna current will be noted. Such
trouble will occur whenever both circuits are tuned to the same
frequency and then too closely coupled or when the two circuits
have slightly different frequencies and are coupled together.
Good amateur practice dictates that the antenna circuit should
be tuned as closely as possible to the closed circuit frequency by
watching the antenna ammeter. Then the coupling is reduced
or the circuit detuned until the antenna meter reads about 75
per cent of its maximum reading under any adjustment. Then
there is little likelihood that the transmitter will suddenly decide
to transmit on a frequency to which the receiver is not tuned.
When the antenna tuning condenser is varied, with such loose
coupling between antenna and tuned circuit, there will be but a
single sharp resonance peak. There will be no danger that the
frequency of transmission will jump from one peak to another,
and thereby clutter up the ether with unintelligent dots and
dashes.
361. Methods of connecting oscillator to antenna.—It is pos-
sible for the inductance and capacity of the antenna-counterpoise
MODULATION 445
system to be part ot the tuned circuit, and if so, the greatest amount
of energy transfer will take place from tube to antenna. Modern
methods, however, involve coupling the antenna to the oscillator
through a “tank circuit,” which the tuned circuit is usually
called. Whatever harmonics exist in this circuit find poor coupling
to the antenna and are inhibited in their voyage from the trans-
mitter to the ether.
When the antenna-counterpoise is series tuned, it is done so to
attain a maximum of current in it. On short waves the antenna
is somewhat larger than one whose natural wavelength is the wave-
length of transmission.
This wavelength is re-
duced by means of the
series condenser.
It is possible to put
power into the antenna Gank, =
by what is known as Voltage Feed
the “voltage feed” Se
method which consists ;
; i Counterpoise
eee cuing ene ng of Fie. 292.—Current feed and voltage feed an-
a wire to some high- henna cirenitae
voltage part of the ;
tube’s oscillatory circuit and the other end of the wire to some high-
voltage part of the antenna. This provides sufficient coupling be-
tween transmitter and antenna to excite the latter. (See Fig. 292.)
362. Feeding power through transmission line.—Oftentimes it
is impossible to put the antenna in a clear location near the trans-
mitting apparatus. It is then possible to feed power to the antenna
through a transmission line, as in Fig. 293, which is at low potential
and not tuned to the frequency at which radiation is to occur.
The resistance of this line is low at this frequency, little current
flows in it, little power is lost in it, and the power is finally put into
the antenna which may be located at some advantageous position.
Transformers at the ends of this line couple it to the transmitter
and to the antenna if the line is longer than 3 wavelength.
363. Modulation.—A transmitter may be designed either for
code or voice transmission. More apparatus is required for the
Antenna
446 OSCILLATORS, TRANSMITTERS, ETC.
latter means of communication. In addition to the oscillator tubes
and amplifiers which may finally feed power into the antenna, there
must be a modulating system. One method is to put the audio
voltage variations into the
grid circuit of the oscillator.
These variations are, then,
superimposed upon the a.-c.
plate current and so the out-
put into the antenna is varied
sciator. at the audio-frequency rate.
The system most fre-
quently used is known as the
Heising or constant-current
system. In it, the plate volt-
a. age to an oscillator, whose
Fic. 293.—Energizing an antenna by
means of a feeder line which does not frequency 18 the baa sr or
radiate. igh frequency, is varied by
the audio-frequency modu-
lating voltages. Since the oscillating current and hence the
antenna current is proportional to the plate voltage (Fig. 280),
this current will vary with the audio variations.
Consider Fig. 294, in which the reactance of the choke L is
high at all audio frequencies compared to the two resistances. The
resistance Rm is the resistance of
what later will be seen to be the
modulator tube, which is simply
a power amplifier operating at
audio frequencies. The resistance
Rk, is the resistance of the escil-
lator tube. Suppose the resistance py¢, 294—Equivalent of Heising
of Rm is caused to vary at some modulation system
audio rate. The current taken
from the B battery will not vary at this rate because of the large
choking effect of the inductance L. The total current, then, from
the battery is constant. If the resistance of the modulator tube,
Rm, increases, less current will be taken by it and more can be
taken by the oscillator tube. On the next half-cycle, the resistance
AMOUNT OF POWER REQUIRED FOR MODULATION 447
of the modulator tube decreases and more current will be taken by
it. The current taken by the oscillator then must decrease.
If the variations in resistance of the modulator tube are at
some audio frequency, say 1000 cycles, the current taken by the
oscillator will vary at this rate too—which is another way of stating
that the r.-f. currents generated by the oscillator and transferred
to the antenna will be modulated at the rate of the audio variations
in the modulator circuit.
The actual circuit is shown in Fig. 295. When the r.-f. output
of the oscillator is completely modulated, it looks as in Fig. 296
Oscillator (Hartley)
Modulator
Speech Input R.F. Choke
]
|
:
:
|
|
|
|
|
|
{
|
| !
\
Fig. 295.—Practical modulated oscillator circuit.
(taken from Heising, Proc. I.R.E., August, 1921), and the power
in the antenna is 1.5 times as great as when no modulation occurs.
The antenna current meter will read the effective value of the cur-
rent, or the square root of the power, and so will increase about
20 per cent when the wave is completely modulated. (The square
root of 1.5 is 1.226, which is 22.6 per cent greater than 1.0.)
364. Amount of power required for modulation.—It used to be
standard practice to use a modulator tube of the same rating as
the oscillator tube. That is, if the oscillator was a 50-watt tube,
the modulator was a 50-watt tube. Nowadays, however, the
modulator has considerably more power than the oscillator, for the
448 OSCILLATORS, TRANSMITTERS, ETC.
following reason. When the tube circuit oscillates a considerable
part of the energy from the plate battery is taken from the plate of
the tube and used up in the tuned circuit or the antenna as the
case may be. In other words a 50-watt tube may actually take
100 watts from the plate battery. Let us suppose it takes 100
watts. Now to completely modulate the oscillator means that once
in each audio cycle the voltage on the plate of the oscillator will
be reduced to zero, or very near it,and once in each cycle the volt-
age on the tube will be doubled. This means that the modulator
Signal Wave Form
oe ||| en ‘I
TO All
vg
| | i ul i
Completely Modulated Antenna
Current
Fic. 296.—A radio frequency wave before and after complete modulation.
tube must draw as much power as the oscillator, but it cannot
because its plate would burn up—it is not oscillating. It must be
biased so that the no-audio-signal input condition leaves the power
wasted on the plate not over 50 watts. When the microphone is
spoken into the steady plate current will change, showing that it
was not acting as a distortionless amplifier.
The solution is to use more modulator tubes so that the com-
bined plate current times the plate voltage is equal in power to
that taken by the oscillator. Even then complete modulation of
the oscillator is impossible because such a condition would imply
that the plate voltage on the modulator would be reduced to zero
at some instants. This is impossible without severe distortion.
DISTORTION AT RECEIVER 449
The result is that modern equipment operates the modulator tubes
at higher voltages than the oscillator.
There is another difficulty in the modulation procedure. The
antenna current under 100 per cent modulation indicates that 50
per cent more power is being supplied. This power must come from
the modulator. In addition to supplying to the oscillator peak
voltages equal to the steady voltage of the oscillator B battery,
the modulator must supply the additional 50 per cent of power.
Since the modulator, acting as an amplifier, cannot be very effi-
cient, a large battery of tubes is required in the modulating
system.
365. Modulation at low power.—Suppose, however, we modulate
the output of a small tube and amplify it. We shall save on
modulation equipment, because we can modulate a 5-watt tube,
say with a 50-watt tube, and when small quantities of power are
to be used efficiency does not matter. The succeeding power
amplifiers, however, must each carry the additional 50 per cent
increase in power, if the wave is to be completely modulated, and
so little has been gained. It is a problem of whether to build a
large modulating equipment acting at audio frequencies or large
radio amplifier equipment.
366. Distortion at receiver due to complete modulation.—
Detectors work according to a square law at low signal inputs, say
up to about 0.2 volt. Because of this square law detection, dis-
tortion results in the receiver. When the wave is completely
modulated, there will be a second harmonic current in the output of
the detector of 25 per cent of the fundamental. With a good
amplifier, and loud speaker, this distortion is evident to a critical
ear.
Let us look at Fig. 297. This represents the relation between
r.-f. input and a.-f. output from a theoretical detector operating
according to a square law. Since modulation can be thought of as
merely changing the r.-f. signal amplitude at an audio rate, let us
suppose an average value of r.-f. signal is 3 volts, and that the wave
is completely modulated. This means that the peak value will be
3 volts varying about an average value of 3 volts. In other words
at one instant the voltage on the input to the detector will be
450 OSCILLATORS, TRANSMITTERS, ETC.
6 volts and at another will be zero volts. The corresponding a.-f.
output will be 36 and 0 volts. This corresponds, as shown in
Fig. 297, to a wave having a maximum value of 36 volts in one
direction, 0 in the other, or a mean value of 18 volts.
Now suppose the modulation is only 50 per cent producing an
r.-f. wave that varies from 1.5 to 4.5 volts, which in turn produces
an audio wave which va-
ries from 20.25 volts in
one direction to 2.25 in an-
other, or an average value
of 9 volts. Inother words,
doubling the modulation
percentage doubles the
average value of the a.-f.
output. Thus the audio-
frequency output from
such a detector may de-
pend upon the square
of the r.-f. input, but it
depends upon the first
power of the modulation.
In Section 186 we de-
termined the percentage
of harmonic distortion
that occurred in an am-
dee ies ey Wier re plifier when it worked
“ta. 297.—In a square law detector the A.F. a
output is proportional to the square of the R.F. hae! ST es
input. istic. Let us apply the
same procedure to the
detector case and see what it brings. Increasing the r.-f. voltage
a given amount produces a different value of a.f. than decreas-
ing the r.f. the same amount does. In other words, distortion
results. The wave form of the output looks like Fig. 298.
The expression for percentage second harmonic is
all nee ae den) =< day
Tine = date
DISTORTION AT RECEIVER 451
and applying it to the two cases of 100 per cent and 50 per cent
modulation we get
For 100 per cent modulation,
! ; 3(36+0)-9 1
distortion = oa Spor 25 per cent;
and
For 50 per cent modulation,
aetoril + (20.25 + 2.25)-9 1
ChetOr ionic aee ee
20.25 — 2.25 8 or 12.5 per cent.
All of this proves that doubling the modulation doubles the per-
centage of distortion. This is the disadvantage of the square law
detector; the distortion 36
arising from its use
prohibits large-percent-
age modulations. Low
degrees of modulation
mean waste of power.
When linear detect-
ors or demodulators are
used, the percentage ,
of modulation at the Fic. 298.—Distortion arising from square law
transmitter can be in- He acem
creased without increase
of distortion taking place in the detector. At the same time the
increase in modulation will increase the detector output. Since
the detector output is roughly proportional to the product of the
modulation and the r.-f. carrier, increasing the modulation has the
same effect as increasing the power of the transmitter, and to
provide the same service at a given distance, increasing the modula-
tion, can be accompanied by a decrease in r.-f. power. Since the dis-
tance, over which two stations interfere (produce beat interference)
. depends upon the carrier power, any decrease in carrier power
effectively increases the number of stations that can operate in a
452 OSCILLATORS, TRANSMITTERS, ETC.
given area and on a given frequency. Since the noise at the
receiver is proportional to the carrier, increasing the modulation
without increasing the carrier improves the signal to noise (static,
etc.) ratio by about four to one, and so the carrier power could be
reduced still more and still maintain the same service. The modern
tendency is toward greater modulation—approaching 100 per cent
on peaks and averaging perhaps 50 per cent—and greater carrier
power as well.
Linear detection has the added advantage that undesired
signals weaker than the desired signals are wiped out by the
latter. (Radio Broadcast November, 1929, F. E. Terman.)
CHAPTER XVIII
ANTENNAS, TRANSMISSION
Now that we have generated oscillations and induced radio-
frequency currents in an antenna, how it is that these currents
convey intelligence to a receiver perhaps thousands of miles away?
What is the nature of this process? What is the nature of the
invisible and often unpredictable medium between transmitter and
receiver?
367. Radiation resistance.—Let us suppose that two wires
about 30 feet long and about a foot apart are coupled to a 40-meter
transmitter through any of the familiar coupling methods, perhaps
by means of a coupling coil as
in Fig. 299. The current into | .
this double-wire system will gi Zz
rise to a maximum when the
A H
varying tuning condenser
eee eee Fic. 299.—Experimental diagram to
C equal to the product of L demonstrate existence of radiation re-
and C of the oscillator. Sup- AiutRnOe.
pose this current is 1 ampere.
Now let us separate the ends of the two wires farther and farther
until finally they are stretched out straight. More and more power
will be required from the plate battery to maintain a constant
value of current in the wires.
At the same time we shall note that the capacity and induc-
tance have changed somewhat so that some minor changes must
be made in the tuning condenser to keep the wires in resonance
with the oscillator—but these changes cannot account for the
greater power required from the battery to maintain the same
current in the wires. In the first place, the changes in capacity and
453
454 ANTENNAS, TRANSMISSION
inductance are balanced out each time by readjusting the tuning
condenser to resonance. Then there is only resistance to impede
the flow of current. Every time the reactances have been balanced
against each other. Clearly the resistance of the wires has in-
creased. At the same time if we installed a small receiver near the
oscillator and kept moving away from it so that the signals picked
up were of constant strength, we should find that the greater the
power taken from the battery and hence the greater the power
into the antenna, the farther away we could hear the signals.
It is apparent that if the current is still 1 ampere but twice
as much power is taken from the battery to produce the 1
ampere, the resistance of the wires has doubled. The useful part
of this resistance is called the radiation resistance of the wires,
which may now be called the antenna. The power that goes into
this resistance is the power that is effective in carrying intelligent
communication from the transmitter to the receiver.
The total resistance of an antenna may be measured by the
same method used to measure the high-frequency resistance of a
coil. (Section 134.) If enough resistance is added to our 40-meter
antenna system to halve the current, this added resistance is equal
to the resistance already there.
If the resistance is measured
at several wavelengths we
shall get a curve similar to
that in Fig. 300 (taken from
the Proceedings of the I.R.E.,
February, 1920, by T. Johnson,
Naval Aircraft Radio). This
resistance is made up of the
ohmic resistance of the wires,
the losses in the dielectric of
the antenna capacity, the loss of energy due to radiation, and other
small losses. An efficient antenna is one which has a very high
radiation resistance and a low resistance of all other sorts. Then
most of the power put into the antenna from the oscillator will be
radiated. This method of measuring antenna resistance is fairly
accurate if very loose coupling and a sensitive meter are used.
Effective Resistance Ohms
iJ
o
Meters Wave Lengt
Fig. 300.—Antenna resistance.
CALCULATION OF THE RECEIVED CURRENT 455
The radiation resistance of most short-wave antennas used
below the fundamental wavelength is of the order of 100 ohms and
it requires a transmitter output of 100 watts to put 1 ampere into
it. The current that goes into an antenna is a variable factor, and
because one antenna has 1 ampere and another 2 amperes in it
it may not mean that 1 is twice as good as the other. The
place in the antenna-ground or antenna-counterpoise system where
the current is read and the physical surroundings of the antennas
may be much more important in determining the effective radia-
tion than the current into it.
368. The radiation field.—The energy which is taken by the
radiation resistance is used in setting up about the antenna a
radiation field. This field consists of both a magnetic field and an
electrostatic field. This radiation field moves away from the
antenna, with a speed equal to the velocity of light, and its strength
at any distance is inversely proportional to the distance.
When the lines of force making up this radiation field cut a
conductor such asa receiving antenna a voltage is induced in this con-
ductor and if amplified and demodulated it becomes the received part
of the communication thrust upon the ether by the transmitter.
This brief and very inadequate explanation of what happens in
the ether does not state how it happens. Most radio engineers,
however, are interested in the result rather than the means and so
wemust be satisfied withthe knowledgethat current into an antenna
produces a radiation field about the antenna, that the intensity
of this field varies inversely with the square of the distance, and
that when this field cuts across a conductor to which is attached a
receiving apparatus a voltage is developed which is the bearer of
the messages sent out at the transmitter.
369. Calculation of the received current.—lIf the antenna at
the transmitter is the type used on shipboard, that is, a high
“ flat-top,” and a ‘“ down lead,” and if the receiving antenna is
similar in construction, the received current is a function of the
transmitting current and the dimensions of the antennas, and the
distance apart is
188 hh, Is
ON
p)
456 ANTENNAS, TRANSMISSION
where h, = the transmitting antenna height;
h, = the receiving antenna height;
I, = the maximum transmitting current;
R = the receiving antenna resistance;
d = the distance apart in meters;
= the wavelength in meters.
This formula shows that, all other conditions remaining the
same, there will be more received current the shorter the wave-
length, but this formula does not show the fact that short waves
are absorbed more readily than long ones. The greater the height
of the transmitter and receiver antenna the greater the received
current. At the same time the absorption is less because of the
fewer objects in the near field of the antenna and so the higher the
antenna the better it is.
This formula does not include Sa variable factors as skip
distance, sky waves, sunset effects or other vagaries of trans-
mission on short or medium waves. It is not useful on amateur
bands.
370. Types of antennas.—Antennas may take a number of
shapes and sizes. Radiating systems in use in high-power long-
Insulators
O C) 5
<—L= L=—>
Fia. 301.—Half-wave vertical Fic. 302.—Folded half-wave
antenna. antenna.
wave stations comprise a very long (one mile or more) flat-top
about 400 feet high, whereas a short-wave station may have an
antenna consisting of a single wire (Fig. 301), vertical or hori-
zontal (Fig. 302), one-half wavelength long. For example, if the
wave radiated is 20 meters, the total length of antenna and counter-
DIRECTIONAL ANTENNAS 457
poise wire will be 10 meters, or about 30 feet. On shipboard the
“down lead” may come from the end of the horizontal part or
from the center. In portable installations an umbrella type is
used, that is, an insulated mast which is held up in the air by
insulated wires which act as the radiating system.
The larger the antenna the more energy it will pick up, either
from desired stations or from other electrical disturbances such as
static. The tendency is to use large antennas for transmitting, so
that a large amount of power can be put into the radiation field,
and small antennas for receiving so that the ratio of signal to static
will be large.
The ordinary antenna used for broadcast reception consists of
a single wire more or less horizontal or vertical as the case may be,
and up to a hundred feet long. Receivers are engineered with an
average antenna in mind and more recent receivers, using
screen-grid tubes, require no antenna at all, but pick up enough
energy from a small wire or loop to obviate the unsightly and dan-
gerous practice of installing wires on the roofs of buildings, on
telephone poles and in trees.
371. Directional antennas.—For broadcasting intelligence over
a wide area an antenna which transmits equally well in all direc-
tions is desirable. Such is the vertical antenna. When a station is
constructed to operate with one other station only, it is a waste of
power to transmit its signals in all directions. Such stations use
directional antennas, that is, radiating systems which transmit
better in one direction, say north-south, than they do in any other.
Even then energy goes out in two directions, north and south, and
the receiving station is, of course, in only one direction from the
transmitter.
The loop is a directional antenna. It receives better in the
direction towards which the narrow dimension points. Its pick-up
ability is something like Fig. 303, in which it is pointed east-west
and picks up very little energy from a north-south direction. In
connection with a sensitive receiver, it may be used te determine
the direction whence the signals come. It is the heart of the
direction-finding stations which are situated along the coasts of
the world. When a ship wants bearings, its signals are picked up
458 ANTENNAS, TRANSMISSION
by the coastal station which determines the position of its receiving
loop which gives the least signal. A compass is attached to the
I (Received
Current)
Fia. 303.—Directional effect of
reception on a loop.
base of the loop and the indicator
then points out the bearing of the
vessel. A receiving operator in an-
other location also swings his loop
on the vessel and thus two bearings
will be obtained. From them the
master of the ship can tell exactly
where he is with regard to the coast
line. The method is shown in Fig. 304.
Other types of antenna give still
greater directional effects, and those
with reflectors behind them (short waves only) will transmit a
narrow beam of signals in only one direction.
” Coast Line
if
Fic. 304.—Method of plotting a ship’s position by obtaining two bearings
from land stations.
372. Inductance and capacity of antennas.—In Sections 136
and 137 means of measuring the capacity and inductance of an
antenna were discussed. The inductance and capacity are not
NATURAL WAVELENGTH OF ANTENNA 459
concentrated as they are in a coil or condenser but are distributed
throughout the structure. The capacity of the antenna wire may
be with respect to ground which serves as one plate of a condenser
with the antenna wires as the other and the air as the dielectric, or
the capacity may be with respect to some other electrical conductor,
usually of the form of a counterpoise which is merely another
antenna, that is as a rule of larger dimensions and much nearer
the ground. The counterpoise is used because of its lower resist-
ance losses than the average ground.
Cmmfd&A
Flying Boat WE
Length in Feet
Fig. 305.—Characteristices of airplane antenna.
The inductance of most attennas is small, of the order of
50 to 100 microhenrys. The capacity of simple receiving
attennas is of the order of 150-300 mfd. (See Fig. 305.)
373. Natural wavelength of antenna.—Because an antenna has
both capacity and inductance it can be made to resonate to a
certain frequency or wavelength. The formula for the wavelength
or frequency to which an antenna resonates is the same as in any
other circuit which has a capacity and inductance, that is,
1
his 27rV LC ;
A vertical wire grounded at the lower end has a natural wave-
length about 4.0 times its physical length. Thus, if the wire is
20 meters long and is grounded, its natural wavelength is about
80 meters. On the other hand, if the wire is not grounded but is
460 ANTENNAS, TRANSMISSION
considered as being two wires, an antenna and counterpoise with
maximum current in the center, the natural wavelength is roughly
twice the length in meters. Thus, an antenna 20 meters long
(30 feet in antenna and 30 feet in counterpoise) will radiate a wave
of 40 meters if not connected to the ground. This antenna is called
the Hertz because it is the type used by that investigator.
The Hertz antenna can also be made to radiate on any of its
harmonics. It is natural wavelength is 40 meters it will radiate at
20 meters or if excited properly by an oscillator, at 10 meters.
374. Loading an antenna.—In case an antenna does not have
the required inductance to bring its wavelength to the desired
value, additional concentrated inductance can be placed in the
down lead and the antenna “loaded” up to the desired wave-
length. Since the coil cannot radiate to any extent and since its
resistance must be added to the resistance of the antenna system,
loading makes the entire system inefficient. The small antenna
gets poor “ hold” on the ether; the higher current causes greater
power loss.
375. Decreasing the wavelength of an antenna.—If the antenna
is too large to tune to the desired wavelength its natural wavelength
can be reduced by placing a condenser in series with it. This
reduces the effective capacity and as a result reduces its wave-
length. The wavelength cannot be reduced below one-half the
natural wavelength. Here again, loss in power in the condenser
must be subtracted from the power that would go into the radia-
tion resistance and so the losses in the condenser must be paid for
in inefficiency. Fortunately, it is possible to build a condenser
whose resistance may be very low at the frequency at which it is
to be worked, and so a series capacity does not add much resistance
to the antenna.
376. Short-wave transmission.—Because of the fact that short
waves were highly absorbed, it was thought for many years that
this portion of the radio-frequency spectrum was worthless. It
was largely for this reason that amateurs were permitted to
operate there. No one else wanted the short waves. Strangely
enough, these short waves, once thought worthless, are now being
fought for by radio communication companies the world over.
FADING 461
The radiation field as it goes out from the antenna is absorbed
by all conductors which exist in its field. According to theory,
the shorter the wavelength the greater the absorption so that for
waves shorter than 100 meters very little energy arrives at a
receiver any distance away. Amateurs, however, discovered that
these waves did arrive at much more distant points than theory
permitted and a new theory had to be developed.
Waves are radiated from the antenna at all angles to the
horizontal. The ground wave which the old theory dealt with
goes near the surface; some radiation leaves the antenna at a
high angle and shoots off into space. Some distance above the
earth is an ionized layer which is a fairly good conductor of elec-
trical disturbances. It therefore reflects a certain amount of the
sky wave which returns to earth and may be received by any
antenna in its path.
The ground wave is soon absorbed. The sky wave does not
come down to earth in the immediate vicinity of the transmitter.
Between the area covered by the ground wave and the sky wave
there is a dead spot known as the skip distance. Signals are not
received there except with the greatest difficulty and with con-
siderable irregularity. In the daytime this skip distance is about
200 miles at 40 meters and 800 miles at 20 meters. Beyond the
skip distance the signals are audible and fall off in intensity until
they are again inaudible.
By properly choosing the frequency for the time of day it is
possible to maintain a continuous communication with another
station at any given distance. In other words, distance, time of
day, and frequency are related. The table at the back of this
book was collected by L. C. Young of the Naval Research Labor-
atory and published in Radio Broadcast.
It is because of the sky wave that amateurs working with less
than 100 watts in an antenna are frequently able to communicate
over several thousand miles on waves below 80 meters. Above
300 meters the skip distance is negligible, the sky wave is not
important, except as noted in Section 377.
377. Fading.—Because the Heaviside layer, as the ionized con-
ducting layer which about 100 miles above the earth is called,
462 ANTENNAS, TRANSMISSION
varies in height and density from moment to moment and from
day to day and season to season, the reflected wave varies in
intensity. This is one cause of fading. Amateurs and other
workers on short waves are the ones who have to contend most
with this phenomenon since it is less effective on longer waves.
It is of some importance on broadcast frequencies but at lower
frequencies becomes of less and less value. When the sky wave
and the ground wave arrive at a receiving station out of phase
with each other, the received signal will be decreased in strength.
This accounts for some fading experienced on broadcast frequencies.
The automatic volume control discussed in Section 308 will
do away with the effects of fading, even at its worst, if the trans-
mitting station is powerful enough to lay down a good field strength
at the receiver. The weakest signal must be 40 DB above the
noise level. Then the volume control will keep this difference
between noise and static by reducing both when the signal comes
up. If the signal gets below the noise or equal to it, the automatic
increase in sensitivity of the receiver will only bring up the noise
with the signal.
378. Comparison of night and day reception.—The shorter the
wavelength the greater the difference between night and day trans-
mission and reception. On long waves there is little difference.
At night signals are somewhat louder. On broadcast frequencies
the difference is marked, especially in winter. Signals can be heard
at night which are inaudible during the day. On short waves the
skip distance becomes much greater at night. Why is this?
During the day the sun pours radiation into our atmosphere
ionizing the particles which constitute it. These ionized particles
absorb radiation of all kinds. Once absorbed, their energy is
lost and they cannot transmit intelligence to distant receiving
stations. At night this ionization ceases, the absorption of radic
waves decreases and signals again reach out. The Heaviside
layer of ionized particles which reflects the sky wave on the fre-
quencies above 3000 ke. is very low in the daytime. The skip dis-
tance is not so great then as it ison a winter night when the Heavi-
side or reflecting layer is high, 100 miles or more above the earth.
379. Static.—Static is part of the noisy background that is
ELIMINATION OF MAN-MADE INTERFERENCE 463
sometimes of sufficient strength to interfere with the reception of
signals. It is caused by natural phenomena, such as discharges of
electricity between clouds at different potential or from clouds to
earth. It is not the same noise that is caused by a leaky trans-
mission line, defective power transformers, wires rubbing in trees
and sparking, or other noises which come under the heading of
man-made static. All of the latter noises can be eliminated.
Static cannot be eliminated but its effects can be reduced.
A receiver that operates from a very small antenna will pick
up less disturbing noise than a large antenna high above the earth
The loop antenna which can be pointed at the desired signal, or
any other directional antenna, will not pick up static or unwanted
signals from other directions and will produce a greater signal to
static ratio. It has the effect of reducing static.
A number of schemes have been devised to reduce the static to
the level of the signal, but none has yet been made that will
eliminate the static without also eliminating the signal. Both are
produced by the same fundamental phenomenon—the charge and
discharge of a condenser. One is produced in a broadcasting sta-
tion by man, the other in the sky by nature.
Static is more bothersome during hot summer weather when
the clouds are highly charged. The more sensitive one’s receiver
the further away the storms can be and still disturb reception.
Man-made static is bothersome at all times of the year and at any
time of the day or night. In the future all man-made static will
be illegal, just as it is now illegal to operate an automobile without
a mufHer.
380. Elimination of man-made interference.—Considerable
advance has been made towards the elimination of radio inter-
ference from all manner of sparking machines. The sparks must
be eliminated or the electric waves they set up must be prevented
from radiating much energy. If the sparks occur in a large system
of wires the energy radiated may be considerable and may ruin
radio reception over a large area. This radiated energy may be
strongest at the natural frequency of the radiating wires but,
because of the high resistance of the spark gap, the disturbance
may cover a very wide band of frequencies.
464 ANTENNAS, TRANSMISSION
Condensers across the place where the spark occurs reduce
both spark and interference; inductances in series with the power
wires leading to the electrical machine creating the disturbance
prevent the radiations from getting into the power lines and hence
from having much of a radiating system. Combinations of induc-
tances and condensers may filter out the radiations by reducing
or preventing sparks, by shunting them into the ground and by
preventing them from
7 Noise Making getting into power lines
60v—~¢ Peerarae (Fig. 306).
H a | To tell whether noise
rE in a receiver comes from
Fia. 306.—Use of noise ‘‘filter.”’ outside or from the ae
ceiver or power equip-
ment itself it is only necessary to disconnect the antenna. If the
noise persists, its origin is within the receiver or power supply.
If it is reduced, it is being picked up by the antenna. A direc-
tional receiver may be used to determine first its general direction
and finally its exact location. Noise in “a.c. operated ”’ or elec-
tric sets may come in over the power wires and then it will be
heard even though the antenna is disconnected.
The complete elimination of interference from true static and
from man-made radio disturbances is one of the few big radio prob-
lems that have not yet been solved.
INDEX
A
“A” battery, 167
“Abac,”’ 129
Acceptor circuit, 137
Adjusting a vacuum tube voltmeter,
344
Air gap, effect on inductance, 72
Alternating current, effective value of,
95
instantaneous value of, 90
maximum, or peak value of, 91
root mean square value of, 96
work done by, 51
Alternating-current circuit, typical,
192
Alternating-current circuits, 89-119
cycle, 51, 89
definitions used in, 89
frequency, 51, 89
period, 51
power in, 117
series, 109
characteristics of, 113
terminology used re, 95
tube tester, 188
tubes, 190
Alternation, 51, 89
Ammeter-voltmeter method of meas-
uring resistance, 33
Ammeters, 30
Ampere, 9, 11
turns, 48
Amplification factor of tube, 175
constant, to measure, 185
general conditions for, 262
meaning of, 177
Amplification factor of tube, 175
calculation of, 254
radio-frequency, 297
purpose of, 297
use of several stages of, 322
Amplifier, See Audio, 228-295
as oscillator, 417
radio-frequency, 296-334
tube, 202-227
Angles, functions of, 92
Antenna, capacity of, 161, 458
counterpoise, 459
inductance of, 161, 458
loading an, 460
wavelength of, 160
decreasing, 460
natural, 459
Antennas, directional, 457
inductance and capacity of, 458
transmission, 453-464
types of, 456
Anti-resonant circuit, 132, 136
Apparatus, power supply, 390-
414
Apparent and real selectivity, 379
Atom, 3
Audio amplifier, coupling tube to
load, in, 281
description of, 228
designer, rules for, 277
effect of leaky condenser in, 250
stray capacities in, 241, 243
equalizing, 257
filtering in, 289
frequency characteristic of resist-
ance, 233
465
466
Audio amplifier, frequency character-
istic of resistance, 233
impedance, 240, 246
inductance load, 240
inverse duplex, 253
loud speaker of, 263
power for, 263
need of, 228
overloading, 212
power, 258
push-pull, 259
quantitative effect of capacities in,
243
reflex, 252
requirements of, 231
screen-grid, 295
transformer in, advantage of, 249
transformer-coupled, 246, 253
transmission unit of, 266
tube as an, 202-227
tuned inductance, 245
uses of tubes in parallel in, 264
Audio amplifiers, 228-265
cascade, 233
comparisons between, 278
design of, 266-295, 275
regeneration in, 286
transformer-coupled, 253
measurements on, 254
volume control in, 278
Autodyne, 364
detuning loss in, 371
Automatic tuning, 380
volume control, 380
Auto-transformer, 69
B
“B” battery, 167
“B-eliminator” curve, 15
Baffles for loud speakers, 386
“Band pass” filters, experiments
with, 374
receivers, 372, 373, 374
INDEX
Band selector, 302
Batteries, 38
Ae 167
Bx 167
HOVE
requirement of plate, 239
Beat note, 357
zero, 357
Beats, phenomenon of, 357
Bel, 266
Bias, ‘‘C” or grid, 173
for alternating-current tubes, 409
for power tubes, 279
in plate current, effect of, 171
means of obtaining, 196
Bridge methods of determining tube
factors, 185
systems, 329
neutralizing, 331
Broadcast frequency tuning coils, 62
By-pass condenser, 86, 104
C
“C” battery, 173
bias, 171, 173, 196, 279, 409
means of obtaining, 196
Calibrating a wavemeter, 152
by clicks, 156
by harmonies, 153
Calibration of a variable inductance,
71
Capacities, effect of stray, 241
in radio-frequency circuit, 304
quantitative circuit, 75
Capacitive circuit, 75
reactance, 102
comparison of, 104
Capacity, 74-88
as a reservoir, 74
centimeter of, 77
combinations of resistance with,106
in a power supply device, 75
of antenna, 161, 458
of coils, distributed, 146
INDEX
Capacity, of condenser, 77
measurement of, 105, 160
on sharpness of resonance, effect of,
304
reactance, 102, 305
tube input, effect of, 304
unit of, 77
Carrier frequency, 336
wave, 300
Cascade amplifiers, 233
Cathode, 191
Cell, battery, 38
common dry, 40
internal resistance of, 41
polarization of, 43
primary, 38
secondary or storage, 38, 40
testing, 42
Cells in parallel, 44
im series, 44
im series-parallel arrangement, 45
Centimeter of capacity, 77
Characteristic curves, dynamic, 206
of receiving tube, 171
slopes of, as tube constants, 181
of screen-grid tubes, 199
Characteristics of a series circuit, 113
Charged bodies, 1
Charges, electrical, 1
laws of, 2
Choke coil, 104
Choke-condenser, 22, 282
connecting, 283
voltage limits on, 285
Circuit, acceptor, 137
anti-resonant, 132
capacitive, 75
energy in, 80
equivalent tube, 177, 209
filter, for rectifier, 401
regulation of, 403
fundamental rectifier, 390
inductive, 75
measuring resistance of, 159
467
Circuit, receiving, 149
rejector, 136
Rice, 329
selectivity of, 139
series resonant, 120
Circuit diagram of receiver, 192
of signal generator, 377
Circuits, alternating-current, 89-119
bridge, 329
neutralizing, 331
coupled, 57
filtering r.-f., 332
more complicated, 29
oscillating, 415, 430
parallel, 25, 27
primary, 49
secondary, 49
series, 24
alternating-current, 109
resonant, 136
short-wave receiver, 365
typical receiving, 162
Coefficient of coupling, 64
Coil factors, 323
Coils, broadcast frequency tuning, 62
coupling of, 63
distributed capacity of, 146
“honey comb,’ 370
inductance, 62
multilayer, 63
properties of, 149-164
resistance of, 144, 163
measurement of, 158
single slide tuning, 70
Comparison of reactances, 104
night and day reception, 462
push-pull and single tube, 293
Comparisons between amplifiers, 278
Compensating for plate current, 194
Condenser, 75
by-pass, 86
capacity of, 77
measurement of, 160
capacity formulas, 85
468
Condenser, charge in, 75
time constant of, 78, 81
conducting plates of, 84
discharged, 78
electricity in, 77
leaky, effect of, 250
Leyden jar, 76
potential energy of, 76
in electrostatic field, 80
tests, 83
Condenser plates, shape of, 380
Condensers, direct-current resistance
in, 81
in alternating-current circuits, 81
in general, 83
in parallel, 87, 88
in series, 87, 88
leakage resistance of, 81
measurements of capacity of, 105,
160
power loss in, 83
properties of, 149-164
sizes of radio, 85
time constant of, 81
tuning, 86
Conductance, 6, 23
mutual, 180
importance of, 181
measurement of, 184
Conducting plates of condenser, 84
Conductivity, 6
Conductors and insulators, 5
Conservation of energy, law of, 54
Coulomb, 9, 77
Counterpoise, 459
Coupling, 63
coefficient of, 64
effect of, 312
in oscillatory circuit, 421
on secondary resistance, of close,
313
short-wave receiver to antenna, 366
to antenna, too close, 444
tube to load, 281
INDEX
Crystal control apparatus, 437
Current, electric, alternating, 51,
89
definition of, 3, 4
detection of, 29
direct, 52, 89
effect of, on circuit, 29
on inductance, 72
effective, 96
in common devices, 10
induced, 49
lagging, 99
leading, 101
measurement of, 30
meters to measure, 10
oscillatory plate, 419, 441
phase relations of, 97
plate, 166, 169, 194
curves, 172, 174
production of, 38-56
received, calculation of, 455
saturation, 169
and voltage in phase, 97
Curve, ‘‘abac,’’ 129
as tube constant, 181
slope of, 16
resonance, 158
width of, 140
response, of receiver, 301
Curve plotting, 13-16
Curves, characteristic, 171, 181, 199,
206
plate current, 172, 174
plate voltage, 174
Cycle, 51, 89
D
DB, 266
use of, 271
Decibel, 266
table of values of, 267
voltage and current ratios of, 268
_ Demodulation, 338
Design of audio amplifiers, 266-
295
_ Designer of amplifier, rules for, 277
Detection, 335-355
example, 351
experiment, 346
problem, 352
conditions for best, 343
in radio-frequency amplifier, 347
of current, 29
of modulated wave, 341
power, 352
Detector action, 350
as a.-c. voltmeter, 343
distortion from square law, 353
grid leak and condenser, 348
plate circuit, 340
power, 352
simple, 338
Detuning loss in autodynes, 371
Dielectric, 77
constant, 84
nature of, 84
of tuning condensers, 86
Differential resistance of tube, 179
Direct-current collector rings, 52
commutator, 52
generator, 52
internal resistance of, 53
open circuit voltage in, 53
plate current and alternating-
voltage, 346
resistance of a tube, 178
Direction-finding stations, 457
Distorting tubes, 335
Distortion, calculation,
221
caused by overloading, 217
due to complete modulation, 449
curved characteristic, 214
positive grid, 216
from square law detector, 353
Distributed capacity of coils, 146
Dynamic characteristic curves, 206
harmonic,
INDEX
469
E
Effective value of alternating voltage
or current, 95
Efficiency, definition of, 56
Electric generator, 49
Electricity, three fundamental effects
of, 30
in condenser, quantity of, 77
frictional, 81
static, 76, 79
Electrodes, 38
Electrolysis, 39
Electrolyte, 38
Electromagnetic field, 80
induction, 49
Electromagnetism, 46
Electromotive force, 10
of cell, 39
unit of, 10
Electrons, 1
diameter of, 3
in vacuum tubes, 166
Electrostatic field, 80
loud speaker, 387
Energy, electrical, 53
in condenser, 79
kinetic, 53
potential, 53, 76, 80
unit of, 79
Engineering tuned radio-frequency
amplifier, 308
voltage divider, 411
Engineers’ shorthand, 11
“Fqualizing,” 257
Equivalent tube circuit, 177, 209
Espenschied, Lloyd, 298
Ether, 3
Exponents, 11
Fading, 461
Farad, 77
470
Faraday’s discovery, 48
importance of, 49
Fidelity of radio receiver, 296, 297
Field, electrical, 4
magnetic, 47
strength, 47
Field intensity, 47
strength, 297
tables, 298
Filament, in vacuum tube, 165
purpose of, 166
rectifiers, typical, 393
thoriated, 190
types of, 189
voltage, effect of, 168
Filaments, operating in series, 191
reactivating thoriated, 190
Filter, 75
band pass, 374
circuits for tube rectifiers, 401
regulation of, 403
Filtering in audio amplifiers, 289
radio-frequency circuits, 332
Flux, 47
density, 47
Frequencies, alternating-current, 51
at high power stations, 52, 89
standard, 156
Frequency, 51, 89
amplifier, intermediate, 357
choice of, 363
carrier, 336
changers, 363
characteristic of resistance ampli-
. fier, 233, 242
doublers, 438
effect of, on inductance, 72
meter, 150
modulated, 336
of series circuit, resonant, 127
side-band, 336
stability of oscillating circuits, 434
Frictional electricity, 81
Functions of angles, 92
INDEX
G
Gain due tube and due coil, 310
Galvonometer, 31
Gaseous rectifiers, 397
characteristics of, 398
Raytheon tube, 397
Gauss, 47
Generator, 38, 45
electric, 49
alternating-current, 51
direct-current, 52
internal resistance of, 53
signal, 376, 377
Goldsmith, A. N., 297
Grid in vacuum tube, 165
bias, 173, 196
obtaining, 429
distortion due to positive, 216
leak and condenser detector, 348
values, effect of, 350
purpose of, 170
space charge, 200
swing, permissible, 216
voltage, 172
Ground wave, 461
H
Harmonie distortion calculation, 221
Harmonics of oscillator tube, 426
Hartley oscillator, 430, 433
Hazeltine’s patent, discussion in, 319-
323
Heater types of tube, 191
Heaviside layer, 461, 462
| Heising modulation system, 446
Henry, 62
High-frequency response in ampli-
fiers, 245
Hum output, 407
Impedance, 106
amplifier, 240
general expressions for, 107
INDEX 471
Impedance, of parallel circuit, 116 Internal resistance, of tube, 179
of tube, 179 testing, 42
te transformer characteristic, | IR drops, 21, 22
Induced current, 49 J
voltages, 49 Joule, 79
Inductance, 57-73 K
calibration of variable, 71 Keying a transmitter, 443
combinations of, with resistance,
106 MG
effect of air gap on, 72 Lagging current, 99
current on, 72 Leading current, 101
frequency on, 72 Leakage inductance, 67
on sharpness of resonance, 143 lines, 67
leakage, 67 resistance, 81
load amplifier, 240 Lenz’s law, 58
magnitude of, 60 Leyden jar, 76
measurement of, 65, 66 Lines of force, 4
mutual, 63 s magnetic, 47
of antenna, 161, 458 | Long-wave receivers, 369
of coil, 61, 74 Long waves, poor quality on, 372
tuning an, 305 Losses, 328
unit of, 62 Loud speaker, 383
variation of, 72, 73 dynamic, 386
Inductances, typical, 62 baffles for, 386
variable, 71 electrostatic, 387
Inductive circuit, 75 horn type, 384
reactance, 100 i measurements, 388
comparison of, 104 moving coil, 385
Inductors, 62 power for, 263
variable, 71 Lumped voltage on a tube, 182
Tnertia-inductance, 59
Input resistance, effect of negative, M
306 Magnetic field, 47
Instantaneous value of alternating- lines of force, 47
current, 90, 91 strength, 47
means of expressing, 93 Magnetism, 45
Insulators and conductors, 5. Magnets, 45
Intermediate-frequency amplifier, 357 laws of, 45
choice of, 363 permanent, 48
Internal resistance, effect of polariza- | Magnitude of amplified voltage, 208
tion on, 43 of induced voltage, 61
of cell, 41 of inductance, 60, 61
of direct-current generator, 53 of mutual inductance, 63
AT2
Master oscillator systems, 436
Mathematics in study of radio, 13
Maximum oscillatory plate circuit,
419
value of alternating current, 91
Measurement of current, 30
Measurements of capacities, 105
loud speaker, 388
on radio receivers, 375
resistance, 35, 184, 187
Meters to measure current, 10, 29, 30
frequency, 50
voltages, 10
wavelength, 150
Weston model, 32
Microfarad, 85
Micro-microfarad, 85
Milliameter, in measuring resistance,
34
Mistreatment of tubes, 200
Modern receivers, 377
Modulated wave, detection of, 340
Modulation, 336
at low power, 449
distortion at receiver due to, 449
grid-circuit, 337
Heising system, 446
in transmitter, 445
percentage, 337
power required for, 447
Molecular motion, effect on resist-
ance, 8
Moving coil speaker, 385
Multiplier, 32
N
Negative input resistance, 306
Neutralizing bridge circuits, 331
Neutrodyne, 330
O
Obtaining grid bias, 429
Oersted’s experiment, 45
Ohm. 7
INDEX
Ohm’s law, 20-37
graphs of, 22-24
ways of stating, 20
Operating filaments in series, 191
Operating points, 202
Oscillating circuits, 415
coupling in, 421, 444
dynamic characteristics of, 422
frequency doublers, 438
highly damped, 416
practical, 430
stability of, 434
various, 432
Oscillation, conditions for, 418, 423
in radio-frequency amplifiers, 325
losses due to, 328
Oscillations, undamped or continu-
ous, 417
Oscillator, adjusting, 433
amplifier of, 417
connecting to antenna, 444
crystal control in, 4387
efficiency of, 425
Hartley, 430, 433
maximum power output of, 429
shunt feeding, 431
systems, master, 486
tube, harmonies of, 426
power output of, 427
transmitters, etc., 415-452
Output choke, 22
devices, 281
transformer, 280
coupling tube of, to load,
281
loss in, 280
turns ratio in, 281
Overall voltage amplification, 238
calculation of, 254
Overloading, amplifier, 212
P
Parallel, cells in, 44
tubes in, 264
INDEX 473
Parallel circuits, 25, 115
characteristics of, 27
impedance of, 116
phase in, 116
resonance, 131, 136
Peak value of alternating current, 91
Pentode, 226
Period, 51
Permeability of iron, 47
of core, 60
Permissible grid swing, 216
Phase, 91
angle, 91, 94
sine of, 91, 92
in parallel circuits, 116
in series circuit, 111
of H,, H,, and Ig, 207
relations between current and volt-
age, 97
Plate, battery requirements, 239
circuit detector, 340
current, compensating for, 194
curves, 172, 174
direct-current, as function of
alternating-current voltage,
346
oscillatory, 419
symbol, 166
oscillator, 441
in vacuum tube, 165, 184
purpose of, 166
resistance, measurement, 187
voltage, effect of, 168, 170, 174
Polarization, 43
means to overcome effects of, 43
Poles, 47
Power amplification, 211, 262
amplifier, 258
apparent, 119
detection, 352
diagrams, 222
effective, 119
electrical, 53
definition of, 53
Power expressions for, 55
factor, 119
feeding, through transmission line,
445
for loud speaker, 263
in alternating-current circuits, 117
in transformer circuits, 68
into resonance circuit, 126
loss in condensers, 83
lost in resistance, 54
output, 210
calculation, 221
of oscillator tube, 427
supply apparatus, 390-414
Protons, 1
Push-pull amplifier, 259
tube, comparison of, 293
R
Radiation field, 455
resistance, 453
Radio-frequency amplifier, detection
in, 347
engineering tuned, 308
intermediate-frequency, 357
losses, 328
negative input resistance in, 306
Neutrodyne, 330
purpose of, 299
phenomena, summary of, 319
task of, 300
tube input capacity of, 304
Radio-frequency amplifiers, 296-334
in general, 303
neutralizing bridge circuits in, 331
regeneration and oscillation in, 325
selectivity in, 315-319
screen-grid tubes as, 334
tunea, 305
Radio-frequency receiving systems,
301, 356
Radiola 60 series, 359, 363, 373
Ratio, arms, 37
turns, 281
474
Ratio, voltage and current, 268
Raytheon tube, 397
Reactance, capacitive, 102
capacity, 102
comparison of inductive and capaci-
tive, 104
inductive, 100
Reactivating thoriated filaments, 190
Receiver, ‘‘band pass,” 372, 373, 374
broadcast, 300
current diagram of, 192
coil factors in, 323
distortion at, 449
fidelity of, 296
long-wave, 369
measurements on, 375
modern, 377
Radiola 60 series, 359, 363, 373
response curve of, 301
selectivity of, 296
sensitivity of, 296
shielded, 381
“short-wave,” 364
Sparton, 374
superheterodyne, 356, 358
telephone, 387
tuning a, 149
Receiving circuits, typical, 162
Receiving systems, 356-389
types of, 301, 356
Reception, night and day, 462
Rectifier, 75
circuit, fundamental, 390
-filter system, 404
gaseous, 397
characteristics of, 398
Raytheon tube, 397
kinds of, 392
single-wave, 395
sulphide, 401
tubes, requirements for, 394
Tungar, 399
typical filament, 393
filter circuit for, 401
INDEX
Rectifiers and power supply appara-
tus, 390-414
Reflex amplifiers, 252
system, 252
Regeneration in audio amplifiers, 286
in radio-frequency amplifiers, 325
Regulation of filter circuit, 403
Rejector circuit, 136
“Repeat points,” 360
Resistance, 6
box, 37
coil, bearing on selectivity, 163
measurement of, 158
coils, 144
combinations of, 106
differential, 179
direct-current of tube, 178
effect of molecular motion on, 8
of temperature on, 8
on series resonant circuit, 125
effective, 134
high-frequency, 45
internal, of cell, 41
of generator, 53
of tube, 179
leak, obtaining grid bias by, 429
leakage, 81
measurement, 33-37, 184, 187
negative input, 306
output load, 205
plate, 179, 184, 187
radiation, 453
ratio arms, 37
temperate, coefficient of, 8
unit of, 7
Resonance, 114, 120-148
circuit, 120
power into, 126
curve, width of, 140
parallel, 131
sharpness of, 138, 140
Resonant frequency of circuit, 127,
134
Response of receiver, 301
INDEX
Rice circuit, 329
Root mean square value, 96
S
Saturation current, 169
Screen-grid tube, 197
at short waves, 367
audio amplifier, 295
characteristic curves of, 199
as radio-frequency amplifier, 334
Selectivity, 315
apparent and real, 379
of circuit, 139
combinations, 274
of radio receiver, 296, 301
of superheterodyne, 363
to signals far off resonance, 322
Self-inductance, 59
Sensitivity of meters, 32
of radio receiver, 296
Series, aiding, 63, 64
cells in, 44
operating filaments in, 191
opposing, 64
Series, alternating-current circuits,
109
characteristics of, 113
phase in, 111
resonant, 120
Series and parallel circuits, 24
resonant, 136
acceptor, 137
rejector, 136
uses of, 136
Series resonant circuit, 120
characteristic of, 123
effect of resistance on, 125
power into, 126
resonant frequency of, 127
Shape of condenser plates, 380
Sharpness of resonance, 138
effect of inductance and capacity
on, 143
475
Short-wave receiver,
antenna, 366
Short-wave receiver circuits, 365
transmission, 460
“Short-wave”’ receivers, 364
Short waves, screen-grid tube at, 367
Shunt-feeding oscillators, 431
Side band, 300
cutting, 301
frequency, 336
Signal generator, 376, 377
Sine wave of voltage, 51
Single slide tuning coil, 70, 71
Single-wave rectifier, 395
Skip distance, 461
Sky wave, 461
Shde wire bridge, 37
Slopes as tube constants, 181
Solenoid, 46, 63
Space charge, 166
grid, 200
“Static,” 83, 462
man-made, 463
elimination of, 463
natural, 463
Static electricity, 76, 79
Sulphide rectifier, 401
Superheterodyne, 356
design, 358
“repeat points” in, 360
selectivity of, 363
Symbols, 17-19
coupling to
ay
Telephone receiver, 387
Temperature, effect on resistance, 8
Temperature coefficient of resistance,
8
Thoriated filaments, reactivating, 190
Time of charge of condenser, 78
constant, 81
Transformer, 66
advantage of, in amplifier, 249
476
Transformer, auto-, 69
characteristics, individual, 291
circuits, power in, 68
-coupled amplifier, 246
losses, 68, 280
output, 280
turns ratio in, 281
with no secondary load, 248
working out of high impedance,
275
Transformers, Amertran DeLuxe,
ratios of, 276
(See Regeneration, 286)
Sangamo Type A, 276
Transmission, 453-464
line, feeding power through, 445
short-wave, 460
unit, 266
Transmitter, field strength of, 297
keying, 443
modulation in, 445
Transmitters, adjusting plate load in,
441
plate current, 441
self-rectified, 440
Transmitting station, high power at,
299
advantage of, 299
purpose of, 299
Triangle functions, 92
TU, 266
Tube, as an amplifier, 202-227
constants, measurement of, 184
C bias for alternating current, 409
factors, methods of determining,
185
filaments, types of, 189
input capacity, effect of, 304
power output of oscillator,427
Raytheon, 397
rectifier, 394
slopes as, 181
tester, alternating-current, 188
vacuum (see Vacuum)
INDEX
Tuned inductance amplifier, 245
radio-frequency amplifiers, 305, 308
receiving set, 356
Tungar rectifier, 399
Tuning condensers, 86
automatic, 380
receiver, 149
Turns ratio in transformer, 281
into detector tube, 324
U
Uses of series and parallel resonant
circuits, 136
of tubes in, 264
Vv
Vacuum tube, 165-201
(See also Detection, 385-355)
“A” battery, 167
alternating-current, 190
alternating-current tester for, 188
amplification constant, 175, 177,
185
“B” battery, 167
characteristic curves, 171
slopes of, as constants, 181
constants, measurements of, 184
slopes as, 181
construction of, 165
direct-current resistance of, 178
equivalent tube circuit, 177, 209
factors, bridge methods of ‘deter-
mining, 185
filament in, 165
purpose of, 166
thoriated, 190
types of, 189
voltage, 168
filaments, operating in series, 191
grid in, 165
bias, 171, 173, 196
purpose of, 170
voltage, 172
INDEX
Vacuum tube, heater types of, 191
cathode, 191
impedance of, 179
internal resistance of, 179
“lumped” voltage on, 182
Measurement of constants, 184—187
mistreatment of, 200
mutual conductance of, 180
importance of, 181
measurement of, 184
plate in, 165
current compensation, 194
current curves, 172, 174
purpose oi plate, 166
resistance, 179, 184, 187
voltage, 168, 170
voltage curves, 174
resistance of, 179
differential, 179
direct-current, 178
internal, 179
plate, 179, 184, 187
saturation current, 169
screen-grid, 197
characteristic curves, 199°
“space charge” in, 166
grid, 200
uses of, in amplifiers, 264
* variable-mu,’”’ 200
voltmeter, 343
use of, 346
Variable inductances, 71
Variometer, 72
Vector, 94
diagrams, 94
vertical component of, 94
Volt, 10
Voltage, amplification, 262
and current ratios, 268
divider, 21, 407, 411
drop, 21
effect of filament, 168
plate, 170
effective, 96
ATT
Voltage, grid, 172, 346
induced, 49, 61
magnitude of, 61
limits on choke-condenser, 284
lumped, on tube, 182
magnitude of amplified, 208
open-circuit, direct-current gener-
ator, 53
phase relations of, 97
plate battery, 239
plate-current curves, 174
regulation, 412
sine wave of, 51
Voltmeter, 10
adjusting, 344
vacuum tube, 343
use of, 346
Voltmeter method of measuring re-
sistance, 34
low resistance, and milliammeter,
34
Voltmeters, 31
multiplier, 32
sensitive, 32
Volume control, 278
automatic, 380
Ww
Wavelength, 129
of antenna, 160, 459, 460
Wavelengths, table of, 370
Wavemeter, 150
calibrating, 152
by clicks, 156
by harmonies, 153
heterodyne, 151
Weston Model meters, 32, 34
Wheatstone bridge, 35, 65
Width of resonance curve, 140
Z
Zero beat, 357
Zero pciential, 83
7 ‘ : a
¢ te y =
7 $<; : > 5
ou : > ; eye 1 a bi ;
i . a,
RELATION BETWEEN Wave Leners in Merers, FREQUENCY IN KILOCYCLES,
AND THE Propuct or InpucTANCE (IN MIcROHENRIES) AND CAPACITY
(In MiIcROFARADS)
Meters| finKe.| LZ XC ||Meters|finKe.| LC ||Meters|finKe.) LXC 5
1 | 300,000/0.0000003]| 450 | 667 | 0.0570 740 | 405 | 0.15415
2 | 150,000/0.0000111|} 460 | 652 .0596 745 | 403 | 0.1562.
8 | 100,000)0.0000018)}| 470 | 639 .0622 750 | 400 | 0.2583
4 | 75,000/0.0000045)]| 480 | 625 .0649 755 | 397 | 0.1604
5 | 60,000)0.0000057|| 490 | 612 .0676 760 | 395 | 0.1626
6 | 50,000}/0.0000101]} 500 | 600 .0704 765 | 392 | 021647
7 | 42,900}/0.0000138]} 505 | 594 .0718 770 | 390 1669
8 | 387,500/0.0000180)]| 510 | 588 .0732 775 | 387 1690
9 | 33,333/0.0000228)| 515 | 583 0747 780 | 385 1712
10 | 30,000/0.0000282)| 520 | 577 0761 785 |. 382 1734
20 | 15,000)0.0001129]| 525 | 572 0776 790 | 380 1756
30 | 10,000)0.0002530]} 530 | 566 0791 795 | 367 1779
40 7,500}0.0004500}| 585 | 561 0806 800 | 375 1801
50 6,000/0.0007040)]} 540 | 556 0821 805 | 373 1824
60 5,000]0.0010140)} 545 | 551 0836 || 810 | 370 1847
70 4,290|0.0018780]] 550 | 546 .0852 815 | 368 1870
80 3,750/0.0018010}} 555 | 541 .0867 820 | 366 1893
90 3,333/0.0022800]} 560 | 536 .0883 825 | 364 1916
100 3,000|0. 00282 565 | 581 .0899 830 | 361 1939
110 2,727|0.00341 570 | 527 0915 835 | 359 1962
120 2,500|0 .00405 575 | 522 0931 840 | 357 1986
130 2,308]0.00476 580 | 517 0947 845+} 355 201
140 2,143]0.00552 os || Gils) 0963 850) | 353 203
150 2,000)0 . 00633 590 | 509 0980 855 | 351 206
160 1,875)0.00721 595 | 504 0996 860 | 349 208
170 1,764)0.00813 600 | 500 1013 865 | 347 21a
180 1,667|0.00912 605 | 496 213
190 1,579)0.01015 610 | 492 1047 875 | 348 216
200 1,500/0.0112 615 } 488
210 1,429)/0.0124 620 | 484
220 1,364/0.01362 625 | 480
(s)
oo
i=)
ee)
va
(=)
oo
ne
or
230 1,304/0.01489 630 | 476 1117 895 | 335 225
240 1,250)0.01621 635 | 472 1135 900 | 333 228
250 1,200)0.01759 640 | 469 1153 905 } 331 231
260 1,154)0.01903 645 | 465 1171 910 | 330 233
270 1,111)0.0205 650 | 462 1189 915 | 328 236
280 1,071)0.0221 655 | 458 1208 920 | 326 238
290 1,034/0.0237 660 | 455 1226 925 | 324 241
300 1,000/0.0258 665 | 451 1245 930 | 323 243
310 968/0.0270 670 | 448 1264 935 | 321 246
320 938]0 .0288 675 | 444 1283 940 | 319 249
330 909}0.0306 680 | 441 1302 945 | 317 251
340 883]0.0325 685 | 438 1321 950 | 316 254
350 857/0. 03845 690 | 435 1340 955 | 314 257
360 834/0.0365 695 | 432 1360 960 | 313 259
370 811|0.0385 700 | 429 1379 965 | 311 262
380 790)0.0406 705 | 426 1399 970 | 309 265
390 769)0 .0428 710 | 423 1419 975 | 308 268
400 750\0.0450 715 | 420 1439 980 | 306 270
410 732|0.0473 720 | 417 . 1459 985 | 305 273
420 715)0.0496 725 } 414 . 1479 990 |. 303 | 0.276
430 698)0 .0520 730 | 411 1500 995 | 302 | 0.279
440 682)0.0545 735 | 408 | 0.1521
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